When and why extraneous solution happen

WEBVTT
Kind: captions
Language: en

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- [Instructor] In other videos,

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we've already introduced the
idea of an extraneous solution

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where you go about solving an equation.

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So you're given an original equation,

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and you do a bunch of algebraic steps,

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and then you solve it, and
you get some solutions.

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And what we've seen, especially when

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we're dealing with radical equations

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and especially some rational equations,

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sometimes all of the
solutions don't work for

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our original equation.

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And so the solutions that don't work,

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that don't work, these are the ones

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that we call extraneous.

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What we're going to do in this video is

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get a little bit deeper.

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We're gonna look at some
extraneous solutions,

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but really think deeply
about why they show up.

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So let's just give an example,

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and this is one that hopefully
you're familiar with,

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or at least the idea of
it you're familiar with,

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when you're dealing
with radical equations.

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Let's say you're trying to
solve two x minus one is equal

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to the square root of eight minus x.

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And I'm gonna go quickly through it.

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If this is unfamiliar to you, I encourage

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you to review solving radical equations.

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But we're gonna really study why

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we get an extraneous solution here.

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So, actually, why don't
you pause this video.

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Figure out what the solutions are.

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I'll give you a hint.

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One of them's going to be extraneous.

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And really think about why

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that extraneous solution shows up.

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All right, now let's do it together

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and just to go through the steps that

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we've seen before for radical equations.

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You wanna square both sides of this.

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So if you square the left-hand side,

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you'll get four x squared, I should say,

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minus four x plus one is equal to,

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you square the square
root of eight minus x.

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You're going to get just eight minus x.

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And now let's see.

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We can subtract eight from both sides,

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and we can add x to both sides,

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so we just have zero
on the right-hand side.

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And on the left-hand side we're going

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to get four x squared minus
three x minus seven is equal

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to zero, so we have an
interesting quadratic here.

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You could use the quadratic formula,

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but it looks like this one
might be factorable by grouping.

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We just have to think of two numbers

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that, when we add them, so a plus b needs

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to be equal to negative three,

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and a times b needs to be equal

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to four times negative seven,

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needs to be equal to negative 28.

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And the obvious ones seem
to be right in front of us,

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four and negative seven.

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Four plus negative seven
is equal to negative three.

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Four times negative seven
is equal to negative 28.

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So we can just break
this negative three up

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into, or negative three x up into a four x

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and a negative seven x.

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So let me do that.

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So you have four x squared,
and then I'm just going to,

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and this is just factoring by grouping.

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Once again, if this is
unfamiliar, I encourage

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you to review that on Khan Academy.

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So we're going to get
plus four x minus seven x,

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and notice these two add
up to negative three x.

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So I'm not changing the equation, really,

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or the value, or what
it's trying to express,

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is equal to zero.

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And in factoring by grouping we can group

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these two together, and then we can group

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these two together.

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And for these first two, we can factor

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out a four x, and so you're going

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to get a four x times x plus one.

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And in these second two we can factor

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out a negative seven, minus
seven times x plus one,

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and all of that is going
to be equal to zero.

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And then we can factor out an x plus one.

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And so we're going to get x plus one

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times four x minus seven.

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So times four x minus
seven is equal to zero.

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And once again, this is just to get us

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to a place where we can look
at what the solutions are.

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So we can see that either
x plus one is going

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to be equal to zero, which happens

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when x is equal to negative one,

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or four x minus seven is equal to zero,

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which happens when four
x is equal to seven,

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which happens when x is equal to 7/4.

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So we've got these two
solutions and I promised you

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that one of them would be extraneous.

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So let's try this first
one, x equals negative one.

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And remember, when you solve
a radical equation like this

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you wanna substitute back
into the original equation

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to see whether they're extraneous or not.

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So when you substitute x
equals negative one back here,

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you're going to get two times negative one

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minus one is going to be
equal to the square root

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of eight minus negative one,

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and so on the left-hand side

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you have negative two minus one.

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That is negative three.

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And you get that equaling
the square root of nine,

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the principal root of nine, and
so you get this weird result

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that negative three is equal to three,

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which we know is not
the case, and that's why

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we know this is extraneous.

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But as I promised, why did this happen?

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And the key is to realize that

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when you're doing algebraic manipulation,

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some operations are reversible,

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and some are not.

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For example, if you say
that a plus b is equal

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to c, and you wanna solve for a,

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and you subtract b from both sides,

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you are going to get a
is equal to c minus b.

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And if you were to start
off with a is equal

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to c minus b, you could add b

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to both sides.

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You could add b to both sides to get back

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to your original equation,

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to get a plus b is equal to c.

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So this adding and subtracting values

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from both sides, that is
a reversible operation.

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You can do it in either direction.

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But when you're squaring things,

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that is not true, and we
did square right over here.

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To go from that first step
to this step, we squared.

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And what we'll see, or what
you might already recognize is

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that squaring is not reversible.

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If I know that a is equal to to b,

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and then I were to square it, we do know

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that a squared is equal to to b squared.

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But if we go the other way around,

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if we know that a squared
is equal to b squared,

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so if we know that a squared
is equal to b squared

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do we know that it's always the case

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that a is equal to b?

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Well I'll give you an example that shows

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that that is not always true.

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We know that negative two squared is equal

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to two squared, but negative
two is not equal to two.

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So that's what we did here.

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We squared at this step,
and it's not reversible,

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and we were able to do a bunch of stuff,

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and we got all the way to the point

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that saying, okay, x is
equal to negative one,

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or x is equal to 7/4, but
it's not always the case

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that when x equals negative one

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that we can reverse all the way back

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to this original equation and
that two x minus one is equal

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to the square root of eight minus x.

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And the two places where this is important

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when you're dealing with algebra,

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the two obvious places
are when you're dealing

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with radical equations,
'cause you're doing

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that squaring step, and that squaring step

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is not reversible.

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The other case is when you are dealing

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with multiplying both sides by a variable.

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And to understand that, if we know

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that a is equal to to b, but then

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we were to multiply both sides
of this equation times x.

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And so we'll get xa is equal to xb.

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That could be true, but is the other way

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around necessarily true?

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If we know that xa is equal to xb,

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is it always the case
that a is equal to b?

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Well, I'll give you an example where

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that isn't the case.

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What happens when x is equal to zero?

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We know that zero times three
is equal to zero times four,

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but it is not the case that
three is equal to four.

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And so this is another situation

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where multiplying both
sides of an equation

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by a variable, you can
go in that direction,

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but it's not necessarily reversible.

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You can't necessarily go
back in that direction.

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And that's why when you solve an equation,

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oftentimes a rational equation,
by multiplying both sides

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of that equation by a
variable and you get a result,

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it's not necessarily
true that you could take

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that result and reverse everything

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and get back to your original equation.

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And that's why you have to
check your extraneous solutions,

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especially for radical equations,

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and especially when you
do an algebraic operation

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like multiplying both sides of an equation

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by a variable or by an expression

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that involves a variable.


Zeros of polynomials (multiplicity)

WEBVTT
Kind: captions
Language: en

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- All right, now let's
work through this together.

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And we can see that all of the choices

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are expressed as a
polynomial in factored form.

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And factored form is
useful when we're thinking

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about the roots of a polynomial,

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the x-values that make that
polynomial equal to zero.

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The roots are also evident when
we look at this graph here.

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We have a root at x equals -4,

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a root at x equals -1.5, or -3/2,

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and a root at x is equal to 1.

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So really what we have to do is say,

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"Which of these factors are consistent

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with the roots that we see?"

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So let's go root by root.

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So here on the left we
have a root at x equals -4.

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In order for this polynomial to be zero

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when x is equal to -4,
that means that x + 4

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must be a factor, or some multiple,

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or some constant times x + 4,

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must be a factor of our polynomial.

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Now we can see in the choices that

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we have a bunch of x + 4s,

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but they have different exponents on them.

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The first one has a 2 as an
exponent, it's being squared,

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while the others have a 1 as the exponent.

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Now what we've talked
about in other videos,

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when we talked about multiplicity,

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we said, "Hey, if we see a sign change

00:01:15.190 --> 00:01:17.770
around a root, like we're
seeing right over here

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around x equals -4,"

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that means that we are going to see

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an odd exponent on the
corresponding factor.

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But if we didn't see a sign change,

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as we see in this other root over here,

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that means we would see an even exponent.

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Now, we clearly see the sign change,

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so we would expect an odd exponent,

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and, of course, 1 is an
odd number and 2 isn't.

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So if you just have a straight up x + 4,

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you would have a sign
change around x equals -4.

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So I can rule out this first choice,

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these other three choices
are still looking good

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based on just the x + 4 factor.

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Now lets move on to the
next factor right over here,

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so, or the next root.

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The next root is at x is equal to -3/2,

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and so one way to think
about it is you could have

00:02:06.290 --> 00:02:10.800
a factor that looks like x + 3/2,

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or this times some constant.

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Now, when we look at the choices,
or the remaining choices,

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we don't see x + 3/2,
but we do see something

00:02:18.980 --> 00:02:21.800
that involves a 2 and a 3, and
so one way to think about it

00:02:21.800 --> 00:02:24.942
is just, "Hey, if I just
multiply this by the constant 2,

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that would get us 2x +
3," well I do see that

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right over here, and
then the next question is

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what should be the exponent?

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Well, once again we have a sign change

00:02:35.190 --> 00:02:38.850
around x equals -3/2, so we would expect

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an odd exponent there.

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And you can see out of the choices,

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only two of them have an exponent of 1,

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which is an odd number,
while the other one

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has an even exponent there, so we can rule

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that one out as well.

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And then we go to this last root.

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Ah, we'll do this in an orange color.

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We have a root at x equals 1,

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so we would expect x - 1, or this

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multiplied by some constant,
to be one of the factors.

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And what's interesting here
is we don't see a sign change

00:03:06.540 --> 00:03:10.680
around x equals 1, so we
would expect an even exponent.

00:03:10.680 --> 00:03:12.980
And so, out of the remaining choices,

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we see an x - 1 in both of them,

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but only choice C has the even exponent

00:03:20.120 --> 00:03:25.120
that we would expect, so
choice C is looking good.

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If we were to look at choice D,

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where this is to the first power,

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we would expect a sign change

00:03:30.970 --> 00:03:33.480
around x is equal to 1, so this would be

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a situation where the curve
would keep going down,

00:03:36.460 --> 00:03:39.050
something like that, so we like choice C.

00:03:39.050 --> 00:03:41.790
Let's do another example.

00:03:41.790 --> 00:03:43.207
So, once again, we are asked,

00:03:43.207 --> 00:03:44.600
"What could be the equation of p?"

00:03:44.600 --> 00:03:46.050
and we're given a graph, so again

00:03:46.050 --> 00:03:48.690
pause this video and try
to work through that.

00:03:48.690 --> 00:03:50.940
All right, we're going
to do the same idea.

00:03:50.940 --> 00:03:53.220
Let's go to this first
root right over here.

00:03:53.220 --> 00:03:56.300
We have a root at x equals -3,

00:03:56.300 --> 00:03:58.780
so we would expect some multiple

00:03:58.780 --> 00:04:02.970
of x + 3 to be one of the factors.

00:04:02.970 --> 00:04:06.910
We also have a sign
change around x equals -3,

00:04:06.910 --> 00:04:11.100
so we would expect an odd multiplicity

00:04:11.100 --> 00:04:13.500
and we would expect an odd exponent

00:04:13.500 --> 00:04:15.420
on the x + 3 factor.

00:04:15.420 --> 00:04:17.020
When we look at all the choices,

00:04:18.000 --> 00:04:18.833
C and D have an even exponent,

00:04:18.833 --> 00:04:20.820
so if we had x + 3 to the fourth,

00:04:20.820 --> 00:04:22.170
then you wouldn't have a sign change here,

00:04:22.170 --> 00:04:23.940
you would just touch the x-axis

00:04:23.940 --> 00:04:26.040
and then go back to
where it was coming from.

00:04:26.040 --> 00:04:28.110
So we can rule out these choices.

00:04:28.110 --> 00:04:30.805
And now, let's look at the second root.

00:04:30.805 --> 00:04:34.710
Right over here, at x is equal to 2,

00:04:34.710 --> 00:04:38.090
so we would expect x - 2
to be one of the factors,

00:04:38.090 --> 00:04:40.500
or a multiple of this,
and because we don't have

00:04:40.500 --> 00:04:42.750
a sign change around x equals 2,

00:04:42.750 --> 00:04:44.590
the graph just touches the x-axis

00:04:44.590 --> 00:04:46.490
and goes back to where it was coming from,

00:04:46.490 --> 00:04:48.950
we would expect an even exponent here.

00:04:48.950 --> 00:04:52.700
And so when we look at the
choices around the x -2 factor,

00:04:52.700 --> 00:04:55.190
we see only one of them
has an even exponent,

00:04:55.190 --> 00:04:58.250
so I am liking Choice B,

00:04:58.250 --> 00:04:59.433
and we are done.


Finding inverses of rational functions

WEBVTT
Kind: captions
Language: en

00:00:00.450 --> 00:00:02.028
- [Instructor] All right,
let's say that we have

00:00:02.028 --> 00:00:07.028
the function f of x and it's
equal to two x plus five,

00:00:07.400 --> 00:00:10.650
over four minus three x.

00:00:10.650 --> 00:00:12.280
And what we wanna do is figure out what

00:00:12.280 --> 00:00:14.310
is the inverse of our function.

00:00:14.310 --> 00:00:15.459
Pause this video and try to figure

00:00:15.459 --> 00:00:17.923
that out before we work on that together.

00:00:18.760 --> 00:00:21.220
All right, now let's work on it together.

00:00:21.220 --> 00:00:22.612
Just as a reminder of what a function

00:00:22.612 --> 00:00:25.260
and an inverse even does,

00:00:25.260 --> 00:00:27.159
if this is the domain of a function

00:00:27.159 --> 00:00:30.490
and that's the set of all
values that you could input

00:00:30.490 --> 00:00:34.370
into the function for x
and get a valid output.

00:00:34.370 --> 00:00:35.760
And so let's say you have an x here,

00:00:35.760 --> 00:00:37.460
it's a member of the domain.

00:00:37.460 --> 00:00:40.700
And if I were to apply the function to it,

00:00:40.700 --> 00:00:42.919
or if I were to input
that x into that function.

00:00:42.919 --> 00:00:47.420
Then the function is
going to output a value

00:00:47.420 --> 00:00:52.420
in the range of the function
and we call that value f of x.

00:00:52.890 --> 00:00:56.180
Now an inverse, that goes the other way.

00:00:56.180 --> 00:01:00.790
If you were to input the f
of x value into the function

00:01:00.790 --> 00:01:03.910
that's going to take us back to x.

00:01:03.910 --> 00:01:07.870
So that's exactly what f inverse does.

00:01:07.870 --> 00:01:09.800
Now how do we actually
figure out the inverse

00:01:09.800 --> 00:01:10.829
of a function especially a function

00:01:10.829 --> 00:01:15.410
that's defined with a
rational expression like this.

00:01:15.410 --> 00:01:17.750
Well the way that I think about it is,

00:01:17.750 --> 00:01:21.440
let's say that y is equal
to our function of x

00:01:21.440 --> 00:01:23.537
or y is a function of x so we could say

00:01:23.537 --> 00:01:27.550
that y is equal to two x plus five,

00:01:27.550 --> 00:01:30.670
over four minus three x.

00:01:30.670 --> 00:01:32.263
For our inverse the relationship

00:01:32.263 --> 00:01:35.200
between x and y is going to be swapped.

00:01:35.200 --> 00:01:37.344
And so in our inverse it's going

00:01:37.344 --> 00:01:42.344
to be true that x is going to
be equal to two y plus five,

00:01:43.680 --> 00:01:47.060
over four minus three y.

00:01:47.060 --> 00:01:49.200
And then to be able to express this

00:01:49.200 --> 00:01:51.716
as a function of x, to say
that what is y as a function

00:01:51.716 --> 00:01:56.020
of x for our inverse we
now have to solve for y.

00:01:56.020 --> 00:01:57.450
So it's just a little bit of algebra here.

00:01:57.450 --> 00:01:59.940
So let's see if we can do that.

00:01:59.940 --> 00:02:01.123
So the first thing that I would do

00:02:01.123 --> 00:02:06.123
is multiply both sides of this
equation by four minus 3 y.

00:02:07.370 --> 00:02:09.685
If we do that, on the left
hand side we are going

00:02:09.685 --> 00:02:12.870
to get x times each of these terms.

00:02:12.870 --> 00:02:16.680
So we're going to get
four x minus three yx

00:02:17.990 --> 00:02:19.450
and then that's going to be equal to

00:02:19.450 --> 00:02:20.917
on the right hand side,
since we multiplied

00:02:20.917 --> 00:02:22.720
by the denominator here we're just going

00:02:22.720 --> 00:02:24.050
to be left with the numerator.

00:02:24.050 --> 00:02:26.960
It's going to be equal to two y plus five.

00:02:26.960 --> 00:02:27.793
And this could be a
little bit intimidating

00:02:27.793 --> 00:02:30.710
'cause we're seeing xs and
ys, what are we trying to do,

00:02:30.710 --> 00:02:33.920
remember we're trying to solve for y.

00:02:33.920 --> 00:02:36.490
So let's gather all
the y terms on one side

00:02:36.490 --> 00:02:39.030
and all the non-y terms on the other side.

00:02:39.030 --> 00:02:41.910
So let's get rid of this two y here.

00:02:41.910 --> 00:02:44.250
Actually, well I could go either way.

00:02:44.250 --> 00:02:45.440
Let's get rid of this two y here,

00:02:45.440 --> 00:02:49.590
so let's subtract two y from both sides.

00:02:49.590 --> 00:02:53.050
And let's get rid of this four
x from the left hand side,

00:02:53.050 --> 00:02:57.270
so let's subtract four x from both sides.

00:02:57.270 --> 00:02:59.820
And then what're we going to be left with.

00:02:59.820 --> 00:03:03.156
On the left hand side
we're left with minus

00:03:03.156 --> 00:03:07.600
or negative five, or actually
it would be this way,

00:03:07.600 --> 00:03:12.600
it would be negative three yx minus two y.

00:03:12.630 --> 00:03:13.538
And you might say hey where is this going,

00:03:13.538 --> 00:03:15.490
but I'll show you in a second,

00:03:15.490 --> 00:03:17.077
is equal to, those cancel out

00:03:17.077 --> 00:03:21.350
and we're gonna have five minus four x.

00:03:21.350 --> 00:03:24.290
Now once again we are
trying to solve for y.

00:03:24.290 --> 00:03:26.680
So let's factor out a y here,

00:03:26.680 --> 00:03:29.670
and then we are going to have y,

00:03:29.670 --> 00:03:33.410
times negative three x minus two

00:03:33.410 --> 00:03:36.460
is equal to five minus four x.

00:03:36.460 --> 00:03:38.560
And now this is the homestretch.

00:03:38.560 --> 00:03:40.684
We can just divide both
sides of this equation

00:03:40.684 --> 00:03:43.206
by negative three x minus two

00:03:43.206 --> 00:03:46.540
and we're going to get y is equal

00:03:46.540 --> 00:03:51.540
to five minus four x, over
negative three x minus two.

00:03:53.200 --> 00:03:54.606
Now another way that
you could express this

00:03:54.606 --> 00:03:55.873
is you could multiply both the numerator

00:03:55.873 --> 00:03:57.870
and the denominator by negative one,

00:03:57.870 --> 00:03:59.170
that won't change the value.

00:03:59.170 --> 00:04:01.170
And then you would get, you would get

00:04:01.170 --> 00:04:03.210
in the numerator four x minus five,

00:04:03.210 --> 00:04:06.860
and in the denominator you
would get a three x plus two.

00:04:06.860 --> 00:04:07.990
So there you have it.

00:04:07.990 --> 00:04:10.900
Our f inverse as a function of x,

00:04:10.900 --> 00:04:12.750
which we could say is equal to this y

00:04:12.750 --> 00:04:15.163
is equal to this right over there.


Rational equations intro

WEBVTT
Kind: captions
Language: en

00:00:00.530 --> 00:00:01.710
- [Instructor] Let's say we wanna solve

00:00:01.710 --> 00:00:03.340
the following equation for x.

00:00:03.340 --> 00:00:08.340
We have x plus one over nine
minus x is equal to 2/3.

00:00:09.440 --> 00:00:11.540
Pause this video and
see if you can try this

00:00:11.540 --> 00:00:13.145
before we work through it together.

00:00:13.145 --> 00:00:16.390
All right now let's work
through this together.

00:00:16.390 --> 00:00:18.530
Now, the first thing
that we might wanna do,

00:00:18.530 --> 00:00:20.150
there's several ways that
you could approach this,

00:00:20.150 --> 00:00:22.140
but the thing I like to do is

00:00:22.140 --> 00:00:24.510
get rid of this x here in the denominator.

00:00:24.510 --> 00:00:26.681
And the easiest way I
can think of doing that,

00:00:26.681 --> 00:00:29.090
is by multiplying both sides

00:00:29.090 --> 00:00:34.050
of this equation by nine minus x.

00:00:34.050 --> 00:00:35.550
Now, when you do that,

00:00:35.550 --> 00:00:38.100
it's important that you
then put the qualifier

00:00:38.100 --> 00:00:40.530
that the x cannot be equal to the value

00:00:40.530 --> 00:00:43.080
that would have made this
denominator zero 'cause clearly

00:00:43.080 --> 00:00:45.650
if somehow you do all this
algebraic manipulation

00:00:45.650 --> 00:00:47.500
and you got x is equal to nine

00:00:47.500 --> 00:00:49.190
that still wouldn't be a valid solution

00:00:49.190 --> 00:00:50.990
'cause if you were to substitute nine

00:00:50.990 --> 00:00:52.730
back into the original equation

00:00:52.730 --> 00:00:55.390
you'd be dividing by
zero in the denominator.

00:00:55.390 --> 00:00:57.050
So, let's just put that right over here,

00:00:57.050 --> 00:01:00.890
x cannot be equal to nine.

00:01:00.890 --> 00:01:03.760
And so then, we can safely move ahead

00:01:03.760 --> 00:01:05.690
with our algebraic manipulations.

00:01:05.690 --> 00:01:07.240
So on the left-hand side,

00:01:07.240 --> 00:01:10.910
as long as x does not
equal nine, if we multiply,

00:01:10.910 --> 00:01:13.480
and divided by nine
minus x they cancel out,

00:01:13.480 --> 00:01:15.950
and we'll just be left with an x plus one,

00:01:15.950 --> 00:01:17.360
and on the right-hand side,

00:01:17.360 --> 00:01:20.080
if you multiply 2/3 times nine minus x,

00:01:20.080 --> 00:01:23.240
we get 2/3 times nine is six

00:01:23.240 --> 00:01:27.447
and then 2/3 times negative
x is negative 2/3 x

00:01:28.460 --> 00:01:30.490
and once again, let's remind ourselves,

00:01:30.490 --> 00:01:33.220
that x cannot be equal to nine.

00:01:33.220 --> 00:01:36.180
And then we can get all of
our x's on the same side

00:01:36.180 --> 00:01:37.390
so let's out that on the left.

00:01:37.390 --> 00:01:39.840
So let's add 2/3 x to both sides.

00:01:39.840 --> 00:01:44.840
So plus 2/3, 2/3 x plus 2/3 x,

00:01:46.600 --> 00:01:49.440
and then, what do we have?

00:01:49.440 --> 00:01:52.320
Well, on the left-hand side we have one x

00:01:52.320 --> 00:01:56.670
which is the same thing
as 3/3 x plus 2/3 x

00:01:56.670 --> 00:02:01.670
is going to give us 5/3 x
plus one is equal to six,

00:02:03.950 --> 00:02:08.120
and then these characters cancel out.

00:02:08.120 --> 00:02:11.490
And then we can just
subtract one from both sides,

00:02:11.490 --> 00:02:16.490
and we get 5/3 x, 5/3 x is equal to five.

00:02:18.810 --> 00:02:20.290
And then, last but not least,

00:02:20.290 --> 00:02:22.310
we can multiply both
sides of this equation,

00:02:22.310 --> 00:02:24.410
times the reciprocal of 5/3

00:02:24.410 --> 00:02:26.220
which is of course 3/5,

00:02:26.220 --> 00:02:27.130
and I'm doing that

00:02:27.130 --> 00:02:30.120
so I just have an x isolated
on the left-hand side.

00:02:30.120 --> 00:02:31.700
So times 3/5,

00:02:31.700 --> 00:02:34.660
and we are left with 3/5 times 5/3

00:02:34.660 --> 00:02:36.560
is of course equal to one.

00:02:36.560 --> 00:02:41.560
So we're left with x is equal
to five times 3/5 is three.

00:02:42.080 --> 00:02:44.040
And so we're feeling pretty
good about x equals three,

00:02:44.040 --> 00:02:46.740
but we have to make sure
that that's consistent

00:02:46.740 --> 00:02:48.780
with our original expression.

00:02:48.780 --> 00:02:49.870
Well if we look up here,

00:02:49.870 --> 00:02:52.220
or if you substitute back x equals three,

00:02:52.220 --> 00:02:53.950
you don't get a zero in the denominator,

00:02:53.950 --> 00:02:55.700
x is not equal to nine.

00:02:55.700 --> 00:02:57.720
X equals three is consistent with that.

00:02:57.720 --> 00:02:59.870
So we should feel good about our solution.

00:02:59.870 --> 00:03:01.430
If we did all this algebraic manipulation

00:03:01.430 --> 00:03:03.250
and we get an x is equal to nine,

00:03:03.250 --> 00:03:05.070
then that still wouldn't
be a valid solution

00:03:05.070 --> 00:03:06.770
because it would have made

00:03:06.770 --> 00:03:10.223
the original expression
on the left be undefined.


Quadratic systems: a line and a parabola

WEBVTT
Kind: captions
Language: en

00:00:00.420 --> 00:00:02.020
- [Instructor] We're
told the parabola given

00:00:02.020 --> 00:00:06.490
by y is equal to three x
squared minus six x plus one

00:00:06.490 --> 00:00:09.550
and the line given by y minus x plus one

00:00:09.550 --> 00:00:11.020
equals zero are graphed.

00:00:11.020 --> 00:00:13.010
So we can see the parabola here in red,

00:00:13.010 --> 00:00:15.000
and we can see the line here in blue.

00:00:15.000 --> 00:00:16.310
And the first thing they ask

00:00:16.310 --> 00:00:20.600
us is one intersection point
is clearly identifiable

00:00:20.600 --> 00:00:21.580
from the graph.

00:00:21.580 --> 00:00:22.413
What is it?

00:00:22.413 --> 00:00:23.570
And they want us to put it in here.

00:00:23.570 --> 00:00:25.330
This is actually a
screenshot from the exercise

00:00:25.330 --> 00:00:26.910
on Khan Academy, but I'm just going

00:00:26.910 --> 00:00:27.743
to write on it.

00:00:27.743 --> 00:00:28.620
If you were doing it on Khan Academy,

00:00:28.620 --> 00:00:29.630
you would type it in.

00:00:29.630 --> 00:00:31.340
But pause this video and
see if you can answer

00:00:31.340 --> 00:00:32.643
this first part.

00:00:33.810 --> 00:00:37.008
All right, so one intersection
point is clearly identifiable

00:00:37.008 --> 00:00:37.950
from the graph.

00:00:37.950 --> 00:00:39.670
I see two intersection points.

00:00:39.670 --> 00:00:42.330
I see that one, and I see that one there.

00:00:42.330 --> 00:00:44.580
This second one seems clearly identifiable

00:00:44.580 --> 00:00:46.364
because, when I look at the grid,

00:00:46.364 --> 00:00:49.350
it looks clearly to be at a value

00:00:49.350 --> 00:00:51.830
of x equals two and y equals one.

00:00:51.830 --> 00:00:54.570
It seems to be the point two comma one.

00:00:54.570 --> 00:00:56.790
So it's two comma one there.

00:00:56.790 --> 00:01:00.000
And what's interesting about
these intersection points is,

00:01:00.000 --> 00:01:02.730
because it's a point
that sits on the graph

00:01:02.730 --> 00:01:06.520
of both of these curves, that means

00:01:06.520 --> 00:01:09.070
that it satisfies both of these equations,

00:01:09.070 --> 00:01:12.520
that it's a solution to
both of these equations.

00:01:12.520 --> 00:01:14.860
So the other one is find the
other intersection point.

00:01:14.860 --> 00:01:16.360
Your answer must be exact.

00:01:16.360 --> 00:01:19.120
So they want us to figure
out this intersection point

00:01:19.120 --> 00:01:20.300
right over here.

00:01:20.300 --> 00:01:22.170
Well to do that, we're going to have

00:01:22.170 --> 00:01:24.730
to try to solve this system of equations,

00:01:24.730 --> 00:01:26.630
and this is interesting
because this is a system

00:01:26.630 --> 00:01:29.590
of equations where one of
the equations is not linear,

00:01:29.590 --> 00:01:31.460
where it is a quadratic.

00:01:31.460 --> 00:01:33.470
So let's see how we could
go about doing that.

00:01:33.470 --> 00:01:34.840
So let me write down the equations.

00:01:34.840 --> 00:01:39.840
I have y is equal to three x
squared minus six x plus one.

00:01:41.040 --> 00:01:42.627
And our next equation right over here,

00:01:42.627 --> 00:01:47.627
y minus x plus one is equal to zero.

00:01:49.650 --> 00:01:51.820
Well one way to tackle,
and this is one way

00:01:51.820 --> 00:01:55.030
to tackle any system of equations,
is through substitution.

00:01:55.030 --> 00:01:58.920
So if I can rewrite
this linear equation as

00:01:58.920 --> 00:02:01.480
in terms of y, if I can solve for y,

00:02:01.480 --> 00:02:03.630
then I can substitute what y equals back

00:02:03.630 --> 00:02:06.750
into my first equation,
into my quadratic one,

00:02:06.750 --> 00:02:09.100
and then hopefully I can solve for x.

00:02:09.100 --> 00:02:10.900
So let's solve for y here.

00:02:10.900 --> 00:02:11.944
And actually, let me color code it,

00:02:11.944 --> 00:02:16.944
because this one is in red,
and this one is the line

00:02:18.340 --> 00:02:20.220
in that blue color.

00:02:20.220 --> 00:02:21.460
So let's just solve for y.

00:02:21.460 --> 00:02:25.510
The easiest way to solve for
y is to add x to both sides

00:02:26.530 --> 00:02:29.220
and subtract one from both sides.

00:02:29.220 --> 00:02:30.800
That was hard to see.

00:02:30.800 --> 00:02:33.370
So and subtract one from both sides.

00:02:33.370 --> 00:02:36.710
And so we are going to get y,

00:02:36.710 --> 00:02:38.630
and then all the rest of
the stuff cancels out,

00:02:38.630 --> 00:02:42.380
is equal to x minus one.

00:02:42.380 --> 00:02:45.380
And so now we can
substitute x minus one back

00:02:45.380 --> 00:02:50.380
in for y, and so we get
x minus one is equal

00:02:50.420 --> 00:02:55.420
to three x squared minus six x plus one.

00:02:56.560 --> 00:02:58.980
Now we wanna get a zero one side

00:02:58.980 --> 00:03:01.581
of this equation, so let's subtract x.

00:03:01.581 --> 00:03:04.020
I'll do this in a neutral color now.

00:03:04.020 --> 00:03:06.530
Let's subtract x from both sides.

00:03:06.530 --> 00:03:09.260
And let's add one to both sides.

00:03:09.260 --> 00:03:11.030
And then whatta we get?

00:03:11.030 --> 00:03:14.110
On the left-hand side we just get zero.

00:03:14.110 --> 00:03:17.320
And on the right-hand side
we get three x squared

00:03:17.320 --> 00:03:20.300
minus seven x plus two.

00:03:20.300 --> 00:03:24.280
So this is equal to zero.

00:03:24.280 --> 00:03:27.530
Now we could try to factor this.

00:03:27.530 --> 00:03:29.950
Let's see, is there an
obvious way to factor it?

00:03:29.950 --> 00:03:33.840
Can I think of two numbers, a times b,

00:03:33.840 --> 00:03:37.070
that's equal to the
product of three and two?

00:03:37.070 --> 00:03:39.200
Three times two, and if
this looks unfamiliar,

00:03:39.200 --> 00:03:41.060
you can review factoring by grouping.

00:03:41.060 --> 00:03:43.470
And can I think of those same two,

00:03:43.470 --> 00:03:45.070
a plus b, where it's going to be equal

00:03:45.070 --> 00:03:46.350
to negative seven?

00:03:46.350 --> 00:03:50.690
And actually, negative
six and negative one work.

00:03:50.690 --> 00:03:54.040
So what can do is I can
rewrite this whole thing

00:03:54.040 --> 00:03:57.230
as zero is equal to three x squared,

00:03:57.230 --> 00:03:59.910
and then instead of negative seven x,

00:03:59.910 --> 00:04:04.180
I can write negative six x and minus x.

00:04:04.180 --> 00:04:05.890
And then I have my plus two.

00:04:05.890 --> 00:04:07.310
I'm just factoring by grouping,

00:04:07.310 --> 00:04:09.150
for those of you are not
familiar with this technique.

00:04:09.150 --> 00:04:11.230
You could also use the quadratic formula.

00:04:11.230 --> 00:04:13.500
So then zero is equal to,

00:04:13.500 --> 00:04:17.680
so if I group these first two I can factor

00:04:17.680 --> 00:04:18.640
out a three x.

00:04:18.640 --> 00:04:23.470
So I'm going to get three
x times x minus two,

00:04:23.470 --> 00:04:28.250
and in these second two I can factor out.

00:04:28.250 --> 00:04:30.920
In these second two I can
factor out a negative one,

00:04:30.920 --> 00:04:34.360
so I have negative one times x minus two.

00:04:34.360 --> 00:04:36.430
And then I can factor out a negative two.

00:04:36.430 --> 00:04:39.030
I'll scroll down a little
bit so I have some space.

00:04:39.030 --> 00:04:42.810
So I have zero is equal to,

00:04:42.810 --> 00:04:45.690
if I factor out an x minus two,

00:04:45.690 --> 00:04:50.690
I'm going to get x minus
two times three x minus one.

00:04:51.740 --> 00:04:54.370
And so a solution would be a situation

00:04:54.370 --> 00:04:56.450
where either of these is equal to zero,

00:04:56.450 --> 00:04:58.650
or, I'll scroll down a little bit more.

00:04:58.650 --> 00:05:01.430
So x minus two could be equal to zero,

00:05:01.430 --> 00:05:04.760
or three x minus one is equal to zero.

00:05:04.760 --> 00:05:06.700
The point at which x
minus two equals zero is

00:05:06.700 --> 00:05:08.660
when x is equal to two.

00:05:08.660 --> 00:05:10.470
And for three x minus one equals zero,

00:05:10.470 --> 00:05:11.510
add one to both sides.

00:05:11.510 --> 00:05:13.310
You get three x is equal to one,

00:05:13.310 --> 00:05:15.440
or x is equal to 1/3.

00:05:15.440 --> 00:05:17.190
So we figured out the,

00:05:17.190 --> 00:05:20.050
we already saw the solution
where x equals two.

00:05:20.050 --> 00:05:21.590
That's this point right over here.

00:05:21.590 --> 00:05:23.200
We already typed that in.

00:05:23.200 --> 00:05:24.570
But now we figured out the x value

00:05:24.570 --> 00:05:25.510
of the other solution.

00:05:25.510 --> 00:05:28.147
So this is x is equal
to 1/3 right over here.

00:05:28.147 --> 00:05:30.230
So our x value is 1/3.

00:05:30.230 --> 00:05:32.210
But we still have to
figure out the y value.

00:05:32.210 --> 00:05:34.300
Well the y value's going
to be the corresponding y

00:05:34.300 --> 00:05:37.670
we get for that x in either equation.

00:05:37.670 --> 00:05:39.960
And I like to focus on the simpler

00:05:39.960 --> 00:05:41.640
of the two equations.

00:05:41.640 --> 00:05:44.760
So we could figure out what is x when,

00:05:44.760 --> 00:05:48.640
or what is y when x is equal
to 1/3 using this equation?

00:05:48.640 --> 00:05:49.590
We could have used the original one,

00:05:49.590 --> 00:05:50.730
but this is even simpler.

00:05:50.730 --> 00:05:52.460
It's already solved for y.

00:05:52.460 --> 00:05:55.620
So y is equal to 1/3 minus one.

00:05:55.620 --> 00:05:58.010
I'm just substituting
that 1/3 back into this.

00:05:58.010 --> 00:06:01.220
And so you get y is equal to negative 2/3.

00:06:01.220 --> 00:06:02.420
And it looks like that as well.

00:06:02.420 --> 00:06:05.580
y is equal to negative
2/3 right over there.

00:06:05.580 --> 00:06:10.500
So is the point 1/3 comma negative 2/3.

00:06:10.500 --> 00:06:11.553
And we're done.


Solving equations by graphing

WEBVTT
Kind: captions
Language: en

00:00:00.140 --> 00:00:02.570
- [Instructor] Let's say you
wanted to solve this equation,

00:00:02.570 --> 00:00:05.666
two to the x squared minus three power

00:00:05.666 --> 00:00:10.666
is equal to one over the cube root of x.

00:00:10.680 --> 00:00:13.823
Pause the video and see
if you can solve this.

00:00:13.823 --> 00:00:17.833
Well you probably realize that
this is not so easy to solve.

00:00:17.833 --> 00:00:20.331
The way that I would at least
attempt to tackle it is,

00:00:20.331 --> 00:00:23.960
you would say this is two
to the x squared minus three

00:00:23.960 --> 00:00:26.990
is equal to x to the,

00:00:26.990 --> 00:00:29.620
I could rewrite this, this
is one over x to the 1/3,

00:00:29.620 --> 00:00:33.030
so this is x to the negative 1/3 power.

00:00:33.030 --> 00:00:35.413
Maybe I can simplify it
by raising both sides

00:00:35.413 --> 00:00:37.803
to the negative three power.

00:00:37.803 --> 00:00:39.206
And so then I would get,

00:00:39.206 --> 00:00:40.453
if I raise something to an exponent,

00:00:40.453 --> 00:00:41.990
then raise that to an exponent,

00:00:41.990 --> 00:00:43.460
I could just multiply the exponents.

00:00:43.460 --> 00:00:46.473
So it would be two to the
negative three x squared

00:00:46.473 --> 00:00:48.190
plus nine power.

00:00:48.190 --> 00:00:51.710
I just multiplied both of these
terms times negative three,

00:00:51.710 --> 00:00:55.260
is equal to x to the negative
1/3 to the negative three.

00:00:55.260 --> 00:00:57.850
Negative 1/3 times
negative three is just 1,

00:00:57.850 --> 00:00:59.582
so that's just going to be equal to x.

00:00:59.582 --> 00:01:04.120
So it looks a little bit
simpler, but still not so easy.

00:01:04.120 --> 00:01:06.460
I could try to take log
base two of both sides

00:01:06.460 --> 00:01:09.587
and I'd get negative
three x squared plus nine

00:01:09.587 --> 00:01:13.190
is equal to log base two of x.

00:01:13.190 --> 00:01:16.650
But once again not havin'
an easy time solve this.

00:01:16.650 --> 00:01:18.660
And the reason why I
gave you this equation

00:01:18.660 --> 00:01:20.439
is to appreciate that some equations

00:01:20.439 --> 00:01:22.807
are not so easy to solve algebraically.

00:01:22.807 --> 00:01:25.995
But we have other tools, we
have things like computers,

00:01:25.995 --> 00:01:28.083
we can graph things and they can

00:01:28.083 --> 00:01:29.450
at least get us really close

00:01:29.450 --> 00:01:32.370
to knowing what the solution is.

00:01:32.370 --> 00:01:33.564
And the way that we can do that

00:01:33.564 --> 00:01:37.156
is we can say hey, well
what if I had one function

00:01:37.156 --> 00:01:42.156
or one equation, that
was y is equal to two x,

00:01:42.260 --> 00:01:45.210
two to the x squared
minus three I should say.

00:01:45.210 --> 00:01:47.697
And you had another that was y is equal

00:01:47.697 --> 00:01:52.697
to one over the cube root of x.

00:01:52.790 --> 00:01:55.280
And then you could graph each of these

00:01:55.280 --> 00:01:57.260
and then you could see
where they intersect.

00:01:57.260 --> 00:01:58.581
Because where they intersect

00:01:58.581 --> 00:02:01.266
that means two to the
x squared minus three

00:02:01.266 --> 00:02:05.170
is giving you the same y as
one over the cube root of x.

00:02:05.170 --> 00:02:06.550
Or another way to think
about it is they're going

00:02:06.550 --> 00:02:08.470
to intersect at an x value,

00:02:08.470 --> 00:02:13.470
where these two expressions
are equal to each other.

00:02:13.950 --> 00:02:15.660
And so what we could do is we could go

00:02:15.660 --> 00:02:17.651
to a graphing calculator,
or we could go to a site

00:02:17.651 --> 00:02:19.960
like Desmos and graph it and at least try

00:02:19.960 --> 00:02:22.890
to approximate what the
point of intersection is,

00:02:22.890 --> 00:02:24.500
and so let's do that.

00:02:24.500 --> 00:02:27.603
So I graphed this ahead of time on Desmos.

00:02:28.520 --> 00:02:32.900
So you can see here, this is
our two sides of our equation

00:02:32.900 --> 00:02:35.390
but now we've expressed
each of them as a function.

00:02:35.390 --> 00:02:39.920
Right here in blue we
have two, we have f of x,

00:02:39.920 --> 00:02:43.358
or I could even say this is
y is equal to f of x which

00:02:43.358 --> 00:02:47.170
is equal to two to the
x squared minus three.

00:02:47.170 --> 00:02:50.620
And then in this yellowish color I have y

00:02:50.620 --> 00:02:52.770
is equal to g of x which is equal

00:02:52.770 --> 00:02:56.090
to one over the cube root of x,

00:02:56.090 --> 00:02:58.410
and we can see where they intersect.

00:02:58.410 --> 00:03:00.417
They intersect right over there.

00:03:00.417 --> 00:03:02.490
And we're not going to
get an exact answer,

00:03:02.490 --> 00:03:04.030
but even at this level of zoom

00:03:04.030 --> 00:03:05.836
and on a tool like Desmos
you can keep zooming in

00:03:05.836 --> 00:03:08.320
in order to get to get a
more and more precise answer.

00:03:08.320 --> 00:03:09.416
In fact you can even scroll over this

00:03:09.416 --> 00:03:11.780
and it'll even tell you
where they intersect.

00:03:11.780 --> 00:03:12.614
But even if we're trying to approximate,

00:03:12.614 --> 00:03:14.860
just looking at the graph.

00:03:14.860 --> 00:03:19.590
We can see that the x
value, right over here

00:03:19.590 --> 00:03:22.800
it looks like it is happening at around,

00:03:22.800 --> 00:03:26.730
let's see this is 1.5, and
each of these is a tenth.

00:03:26.730 --> 00:03:29.480
So this is 1.6 and then it looks like

00:03:29.480 --> 00:03:32.270
it's about two thirds of
the way to the next one.

00:03:32.270 --> 00:03:33.420
So this looks like it's about,

00:03:33.420 --> 00:03:36.833
I'll say this is approximately 1.66.

00:03:39.110 --> 00:03:41.790
And if you were to actually
find the exact solution,

00:03:41.790 --> 00:03:46.320
you would actually find
this awfully close to 1.66.

00:03:46.320 --> 00:03:48.057
So the whole point here is, is that,

00:03:48.057 --> 00:03:50.830
even when it's algebraically difficult

00:03:50.830 --> 00:03:52.570
to solve something you could set up

00:03:52.570 --> 00:03:55.891
or restate your problem
or reframe your problem

00:03:55.891 --> 00:03:58.940
in a way that makes it easier to solve.

00:03:58.940 --> 00:04:02.360
You can set this up as, hey
let's make two functions,

00:04:02.360 --> 00:04:05.030
and then let's graph them
and see where they intersect

00:04:05.030 --> 00:04:06.520
And the x value where they intersect well

00:04:06.520 --> 00:04:08.530
that would be a solution to that equation,

00:04:08.530 --> 00:04:10.660
and that's exactly what
we did right there.

00:04:10.660 --> 00:04:11.748
We're saying that hey, the x value,

00:04:11.748 --> 00:04:16.333
the x solution here is roughly 1.66.


Positive and negative intervals of polynomials

WEBVTT
Kind: captions
Language: en

00:00:00.100 --> 00:00:02.120
- [Instructor] Let's say that
we have the polynomial p of x.

00:00:02.120 --> 00:00:06.170
And when expressed in factored
form, it is x plus two

00:00:06.170 --> 00:00:08.670
times two x minus three

00:00:08.670 --> 00:00:11.210
times x minus four.

00:00:11.210 --> 00:00:13.470
What we're going to do in this
video is use our knowledge

00:00:13.470 --> 00:00:16.330
of the roots of this polynomial
to think about intervals

00:00:16.330 --> 00:00:20.320
where this polynomial would
be positive or negative.

00:00:20.320 --> 00:00:22.820
And the key realization
is, is that the sign

00:00:22.820 --> 00:00:26.600
of a polynomial stays the same
between consecutive zeros.

00:00:26.600 --> 00:00:30.360
Let me just draw an arbitrary
graph of a polynomial here

00:00:30.360 --> 00:00:32.140
to make you appreciate why that is true.

00:00:32.140 --> 00:00:33.890
So x-axis, y-axis.

00:00:33.890 --> 00:00:38.020
And if I were to draw some
arbitrary polynomial like that,

00:00:38.020 --> 00:00:40.120
you can see that between consecutive zeros

00:00:40.120 --> 00:00:40.990
the sign is the same.

00:00:40.990 --> 00:00:44.350
Between this zero and this zero,
the polynomial is positive.

00:00:44.350 --> 00:00:47.540
Between this zero and this zero,
the polynomial is negative.

00:00:47.540 --> 00:00:49.650
And that's almost intuitively true.

00:00:49.650 --> 00:00:52.640
Because if the sign did not stay the same,

00:00:52.640 --> 00:00:54.470
that means you would
have to cross the x-axis.

00:00:54.470 --> 00:00:55.890
So you would have a zero,

00:00:55.890 --> 00:00:59.150
but we're saying between
consecutive zeros.

00:00:59.150 --> 00:01:01.690
So between this zero and this
zero, it is positive again.

00:01:01.690 --> 00:01:03.920
Then after that zero, it stays negative.

00:01:03.920 --> 00:01:06.240
Once again, the only way it
wouldn't stay negative is

00:01:06.240 --> 00:01:08.670
if there were another zero.

00:01:08.670 --> 00:01:10.960
So now let's go back to this example here.

00:01:10.960 --> 00:01:13.380
And let me delete this
because this is not the graph

00:01:13.380 --> 00:01:16.250
of p of x, which I have just written down.

00:01:16.250 --> 00:01:18.260
Let's first think about its zeros.

00:01:18.260 --> 00:01:20.470
So the zeros are the x-values

00:01:20.470 --> 00:01:22.450
that would either make
x plus two equal zero,

00:01:22.450 --> 00:01:25.370
two x minus three equal zero,
or x minus four equal zero.

00:01:25.370 --> 00:01:26.630
So first, we can think about,

00:01:26.630 --> 00:01:29.890
well, what x-values would make x plus two,

00:01:29.890 --> 00:01:32.320
x plus two equal to zero?

00:01:32.320 --> 00:01:35.050
Well, that, of course, would
be x equals negative two.

00:01:35.050 --> 00:01:37.920
What x-values would make two
x minus three equal zero?

00:01:37.920 --> 00:01:40.270
Two x minus three equal to zero,

00:01:40.270 --> 00:01:43.270
add three to both sides,
you get two x equals three.

00:01:43.270 --> 00:01:46.570
Divide both sides by two,
you get x equals 3/2.

00:01:46.570 --> 00:01:50.260
And then last but not least,
what x-values would make

00:01:50.260 --> 00:01:53.320
x minus four equal to zero?
(distant siren wailing)

00:01:53.320 --> 00:01:58.100
Add four to both sides,
you get x is equal to four.

00:01:58.100 --> 00:02:00.840
And so if we were to plot this,

00:02:00.840 --> 00:02:03.860
it would look something like this.

00:02:03.860 --> 00:02:07.129
So this x equals negative two,

00:02:07.129 --> 00:02:09.270
(distant siren wailing)
x equals negative one.

00:02:09.270 --> 00:02:11.280
This is zero.

00:02:11.280 --> 00:02:15.080
This is one, two,

00:02:15.080 --> 00:02:17.970
three, and four.

00:02:17.970 --> 00:02:20.340
And let me draw the y-axis here.

00:02:20.340 --> 00:02:24.990
So the y-axis would look
something like this, x and y.

00:02:24.990 --> 00:02:27.480
We have a zero at x equals negative two,

00:02:27.480 --> 00:02:31.150
so our graph will
intersect the x-axis there.

00:02:31.150 --> 00:02:35.160
We have a zero at x is equal
to 3/2, which is 1 1/2,

00:02:35.160 --> 00:02:36.330
which is right over there.

00:02:36.330 --> 00:02:38.500
And we have a zero at x equals hour,

00:02:38.500 --> 00:02:41.230
which is right over there.

00:02:41.230 --> 00:02:43.930
And so we have several
candidate intervals.

00:02:43.930 --> 00:02:46.610
And actually, let me write
this down in a table.

00:02:46.610 --> 00:02:48.180
So the intervals

00:02:49.370 --> 00:02:50.960
over which and this is really,

00:02:50.960 --> 00:02:54.610
be between consecutive zeros,

00:02:54.610 --> 00:02:56.940
intervals to consider.

00:02:56.940 --> 00:02:59.320
So I'll draw a little table here.

00:02:59.320 --> 00:03:03.680
So you have x is less than negative two.

00:03:03.680 --> 00:03:05.160
That's one interval.

00:03:05.160 --> 00:03:09.340
X is less than, actually,
let me color-code this.

00:03:09.340 --> 00:03:10.890
So if I were to say the interval

00:03:10.890 --> 00:03:13.020
for x is less than negative two,

00:03:13.020 --> 00:03:16.640
so that's this yellow that I
draw at the extreme left there.

00:03:16.640 --> 00:03:18.940
We could have an interval where

00:03:18.940 --> 00:03:22.380
x is between negative two and 3/2,

00:03:22.380 --> 00:03:26.580
so negative two is less than
x, is less than positive 3/2.

00:03:26.580 --> 00:03:30.550
That would be this
interval right over here.

00:03:30.550 --> 00:03:34.720
You have the interval, I'm
trying to use all my colors,

00:03:34.720 --> 00:03:38.570
between 3/2 and four, this interval here.

00:03:38.570 --> 00:03:43.320
So that would be 3/2 is less
than x, is less than four.

00:03:43.320 --> 00:03:45.190
And then last but not least,

00:03:45.190 --> 00:03:48.660
you have the interval where
x is greater than four,

00:03:48.660 --> 00:03:50.450
that interval right over there.

00:03:50.450 --> 00:03:52.870
So x is greater,

00:03:52.870 --> 00:03:55.840
greater than four.

00:03:55.840 --> 00:03:59.810
Now, there's a couple of ways
of thinking about whether,

00:03:59.810 --> 00:04:03.890
over that interval, our function
is positive or negative.

00:04:03.890 --> 00:04:06.603
One method is to just evaluate our p,

00:04:07.558 --> 00:04:10.210
our function at a point in the interval.

00:04:10.210 --> 00:04:11.043
And if it's positive,

00:04:11.043 --> 00:04:12.820
well, that's means that that
whole interval is positive.

00:04:12.820 --> 00:04:13.653
It's negative,

00:04:13.653 --> 00:04:15.390
that means that that whole
interval is negative.

00:04:15.390 --> 00:04:17.760
And once again, it's intuitive because,

00:04:17.760 --> 00:04:19.350
if for whatever reason, it were to switch,

00:04:19.350 --> 00:04:20.740
we would have another zero.

00:04:20.740 --> 00:04:22.470
I know I keep saying that.

00:04:22.470 --> 00:04:24.470
But another way to think about it is,

00:04:24.470 --> 00:04:27.930
over that interval, what is
the behavior of x plus two,

00:04:27.930 --> 00:04:29.970
two x minus three, and x minus four?

00:04:29.970 --> 00:04:31.690
Think about whether they're
positive or negative,

00:04:31.690 --> 00:04:33.990
and use our knowledge
of multiplying positives

00:04:33.990 --> 00:04:36.450
and negatives together to
figure out whether we're dealing

00:04:36.450 --> 00:04:38.210
with a positive or negative.

00:04:38.210 --> 00:04:41.180
So let's do it, we could do it both ways.

00:04:41.180 --> 00:04:44.370
So let's think of this as our sample x,

00:04:44.370 --> 00:04:46.990
sample x-value.

00:04:46.990 --> 00:04:50.040
And then let's see what
we can intuit about

00:04:50.040 --> 00:04:52.520
or deduce about whether,
over that interval,

00:04:52.520 --> 00:04:55.250
we are positive or negative.

00:04:55.250 --> 00:04:58.690
So for x is less than negative two,

00:04:58.690 --> 00:05:01.210
maybe an easy one or
an obvious one to use,

00:05:01.210 --> 00:05:04.230
it could be any value where
x is less than negative two,

00:05:04.230 --> 00:05:08.220
but let's try x is
equal to negative three.

00:05:08.220 --> 00:05:11.650
So you could try to evaluate
p of negative three.

00:05:11.650 --> 00:05:13.140
You could just evaluate that.

00:05:13.140 --> 00:05:14.510
Actually, let's just do that.

00:05:14.510 --> 00:05:19.350
So that's going to be equal
to negative one times,

00:05:19.350 --> 00:05:22.550
two times negative three is negative six,

00:05:22.550 --> 00:05:24.350
minus three is negative nine,

00:05:24.350 --> 00:05:27.510
negative nine times
negative three minus four,

00:05:27.510 --> 00:05:28.693
that is negative seven.

00:05:29.720 --> 00:05:31.540
So if you were to
multiply all of this out,

00:05:31.540 --> 00:05:36.540
this would give you negative
63, which is clearly negative.

00:05:36.580 --> 00:05:39.616
So over this interval right over here,

00:05:39.616 --> 00:05:43.060
our polynomial is going to be negative.

00:05:43.060 --> 00:05:45.410
So then we can move on to the next one.

00:05:45.410 --> 00:05:46.580
An interesting thing
is we didn't even have

00:05:46.580 --> 00:05:47.870
to figure out the 63 part.

00:05:47.870 --> 00:05:49.860
We can just see that there's
a negative times a negative

00:05:49.860 --> 00:05:52.620
times a negative, which
is going to be a negative.

00:05:52.620 --> 00:05:54.300
And so let's just do that going forward.

00:05:54.300 --> 00:05:55.950
Let's just think about whether each

00:05:55.950 --> 00:05:57.650
of these are going to
be positive or negative

00:05:57.650 --> 00:05:58.830
and what would happen when you multiply

00:05:58.830 --> 00:06:01.100
those positive and negatives together.

00:06:01.100 --> 00:06:05.900
Now, in this second interval
between negative two and 3/2,

00:06:05.900 --> 00:06:08.360
what is going to happen?

00:06:08.360 --> 00:06:10.010
Well, we could do a sample point.

00:06:11.310 --> 00:06:12.270
Let's say x is equal to zero.

00:06:12.270 --> 00:06:14.210
That might be pretty straightforward.

00:06:14.210 --> 00:06:16.210
Well, when x is equal to zero,

00:06:16.210 --> 00:06:19.040
we're going to be dealing with a positive

00:06:19.040 --> 00:06:21.920
times a negative times a negative,

00:06:21.920 --> 00:06:24.380
a positive times a
negative times a negative.

00:06:24.380 --> 00:06:25.620
And the reason why I
did that is in my head.

00:06:25.620 --> 00:06:28.540
I said, okay, that's
going to be a positive two

00:06:28.540 --> 00:06:31.090
times a negative three
times a negative four.

00:06:31.090 --> 00:06:33.860
So a positive times a
negative times a negative,

00:06:33.860 --> 00:06:34.730
so I could write it this way.

00:06:34.730 --> 00:06:37.970
It's going to be a
positive times a negative

00:06:37.970 --> 00:06:39.210
times a negative.

00:06:39.210 --> 00:06:41.720
Well, a negative times a
negative is a positive,

00:06:41.720 --> 00:06:43.980
and a positive times a
positive is a positive.

00:06:43.980 --> 00:06:46.130
So we are positive over that interval.

00:06:46.130 --> 00:06:48.210
And if you were to evaluate p of zero,

00:06:48.210 --> 00:06:50.900
you will get a positive value.

00:06:50.900 --> 00:06:52.880
Now what about this next interval?

00:06:52.880 --> 00:06:56.490
What about this next interval
here between 3/2 and four?

00:06:56.490 --> 00:06:59.400
We could try x is equal to two.

00:06:59.400 --> 00:07:01.440
At when x is equal to two,

00:07:01.440 --> 00:07:04.230
we are going to get a positive

00:07:05.100 --> 00:07:07.640
times a positive

00:07:09.040 --> 00:07:13.350
times a, two minus four is
negative, times a negative.

00:07:13.350 --> 00:07:16.960
So this is going to be
negative over that interval.

00:07:16.960 --> 00:07:21.850
And then last but not least,
when x is greater than four,

00:07:21.850 --> 00:07:26.160
we could try x is equal to let's say five.

00:07:26.160 --> 00:07:29.230
We are going to have a positive,

00:07:29.230 --> 00:07:32.710
positive times a positive

00:07:34.180 --> 00:07:36.520
times a positive.

00:07:36.520 --> 00:07:39.670
So we are going to have a positive.

00:07:39.670 --> 00:07:40.720
And as I mentioned,

00:07:40.720 --> 00:07:42.670
you could also do it
without the sample points.

00:07:42.670 --> 00:07:45.230
You could say, okay, when
x is greater than four,

00:07:45.230 --> 00:07:47.030
you could say, okay, for
any x greater than four,

00:07:47.030 --> 00:07:50.110
if you add two to it, that for
sure is going to be positive.

00:07:50.110 --> 00:07:52.340
For any x greater than four,

00:07:52.340 --> 00:07:55.420
if you multiply it by
two and subtract three,

00:07:55.420 --> 00:07:57.730
well, that's still going to be positive

00:07:57.730 --> 00:07:59.380
'cause two times something
greater than four

00:07:59.380 --> 00:08:00.800
is definitely greater than three.

00:08:00.800 --> 00:08:02.430
And for any x greater than four,

00:08:02.430 --> 00:08:03.530
if you subtract four from it,

00:08:03.530 --> 00:08:05.430
you're still going to
have a positive value.

00:08:05.430 --> 00:08:06.750
So that's another way to think about it,

00:08:06.750 --> 00:08:10.030
even if you don't use a sample point.

00:08:10.030 --> 00:08:12.500
But there you have it, we
figured out the intervals

00:08:12.500 --> 00:08:16.330
over which the function
is negative or positive.

00:08:16.330 --> 00:08:19.020
And we don't know exactly
what the function looks like,

00:08:19.020 --> 00:08:20.610
but generally speaking it's negative

00:08:20.610 --> 00:08:21.530
over this first interval.

00:08:21.530 --> 00:08:22.730
So it might look something like this.

00:08:22.730 --> 00:08:25.780
It's positive over that next interval.

00:08:25.780 --> 00:08:28.460
And then it's negative
over that third interval.

00:08:28.460 --> 00:08:30.980
And then it's positive
over that last interval.

00:08:30.980 --> 00:08:32.550
So we'd have a general shape like this.

00:08:32.550 --> 00:08:34.610
We don't know, without
trying out more points,

00:08:34.610 --> 00:08:38.463
exactly how high or low
it would actually go.


Multiplicity of zeros of polynomials

WEBVTT
Kind: captions
Language: en

00:00:00.650 --> 00:00:01.700
- [Instructor] So what we have here

00:00:01.700 --> 00:00:05.100
are two different polynomials, P1 and P2.

00:00:05.100 --> 00:00:07.770
And they have been
expressed in factored form

00:00:07.770 --> 00:00:09.270
and you can also see their graphs.

00:00:09.270 --> 00:00:12.000
This is the graph of Y is
equal to P1 of x in blue,

00:00:12.000 --> 00:00:16.610
and the graph of Y is
equal to P2 x in white.

00:00:16.610 --> 00:00:17.640
What we're going to do in this video

00:00:17.640 --> 00:00:19.410
is continue our study of zeros,

00:00:19.410 --> 00:00:21.000
but we're gonna look at a special case

00:00:21.000 --> 00:00:25.040
when something interesting
happens with the zeros.

00:00:25.040 --> 00:00:28.710
So let's just first look at P1's zeros.

00:00:28.710 --> 00:00:33.060
So I'll set up a little table
here, because it'll be useful.

00:00:33.060 --> 00:00:36.793
So the first column, let's
just make it the zeros,

00:00:37.690 --> 00:00:41.870
the x values at which our
polynomial is equal to zero

00:00:41.870 --> 00:00:44.700
and that's pretty easy to
figure out from factored form.

00:00:44.700 --> 00:00:45.930
When x is equal to one,

00:00:45.930 --> 00:00:47.330
the whole thing's going
to be equal to zero

00:00:47.330 --> 00:00:49.340
because zero times anything is zero.

00:00:49.340 --> 00:00:51.590
When x is equal to two,
by the same argument,

00:00:51.590 --> 00:00:54.060
and when x is equal to three.

00:00:54.060 --> 00:00:55.710
And we can see it here on the graph,

00:00:55.710 --> 00:00:57.330
when x equals one,

00:00:57.330 --> 00:01:00.720
the graph of y is equal to
P1 intersects the x axis.

00:01:00.720 --> 00:01:03.710
It does it again at the
next zero, x equals two.

00:01:03.710 --> 00:01:06.720
And at the next zero, x equals three.

00:01:06.720 --> 00:01:08.210
We can also see the property

00:01:08.210 --> 00:01:11.510
that between consecutive
zeros our function,

00:01:11.510 --> 00:01:14.060
our polynomial maintains the same sign.

00:01:14.060 --> 00:01:15.530
So between these first two,

00:01:15.530 --> 00:01:17.500
or actually before this
first zero it's negative,

00:01:17.500 --> 00:01:19.870
then between these
first two it's positive,

00:01:19.870 --> 00:01:21.270
then the next two it's negative,

00:01:21.270 --> 00:01:23.980
and then after that it is positive.

00:01:23.980 --> 00:01:25.600
Now what about P2?

00:01:25.600 --> 00:01:27.800
Well P2 is interesting,

00:01:27.800 --> 00:01:29.560
'cause if you were to multiply this out,

00:01:29.560 --> 00:01:32.710
it would have the same degree as P1.

00:01:32.710 --> 00:01:36.330
In either case, you would
have an x to the third term,

00:01:36.330 --> 00:01:38.810
you would have a third degree polynomial.

00:01:38.810 --> 00:01:40.080
But how many zeros,

00:01:40.080 --> 00:01:43.130
how many distinct unique
zeros does P2 have?

00:01:43.130 --> 00:01:45.030
Pause this video and think about that.

00:01:45.870 --> 00:01:48.310
Well let's just list them out.

00:01:48.310 --> 00:01:50.713
So our zeros,

00:01:51.730 --> 00:01:54.610
well once again if x equals one,

00:01:54.610 --> 00:01:56.520
this whole expression's
going to be equal to zero,

00:01:56.520 --> 00:01:58.190
so we have zero at x equals one,

00:01:58.190 --> 00:02:00.860
and we can see that our
white graph also intersects

00:02:00.860 --> 00:02:03.110
the x axis at x equals one.

00:02:03.110 --> 00:02:05.800
And then if x is equal to three,

00:02:05.800 --> 00:02:08.250
this whole thing's going
to be equal to zero,

00:02:08.250 --> 00:02:09.970
and we can see that it intersects

00:02:09.970 --> 00:02:12.230
the x axis at x equals three.

00:02:12.230 --> 00:02:16.667
And then notice, this next part
of the expression would say,

00:02:16.667 --> 00:02:18.257
"Oh, whoa we have a
zero at x equals three,"

00:02:18.257 --> 00:02:22.470
but we already said that, so
we actually have two zeros

00:02:22.470 --> 00:02:24.580
for a third degree polynomial,

00:02:24.580 --> 00:02:26.770
so something very
interesting is happening.

00:02:26.770 --> 00:02:28.510
In some ways you could say that hey,

00:02:28.510 --> 00:02:29.730
it's trying to reinforce

00:02:29.730 --> 00:02:31.920
that we have a zero at x minus three.

00:02:31.920 --> 00:02:36.310
And this notion of having multiple parts

00:02:36.310 --> 00:02:40.080
of our factored form that would
all point to the same zero,

00:02:40.080 --> 00:02:43.490
that is the idea of multiplicity.

00:02:43.490 --> 00:02:45.560
So let me write this word down.

00:02:45.560 --> 00:02:47.540
So multiplicity.

00:02:47.540 --> 00:02:49.093
Multiplicity,

00:02:50.830 --> 00:02:52.220
I'll write it out there.

00:02:52.220 --> 00:02:54.723
And I will write it
over here, multiplicity.

00:02:58.360 --> 00:03:01.050
And so for each of these zeros,

00:03:01.050 --> 00:03:03.910
we have a multiplicity of one.

00:03:03.910 --> 00:03:07.220
There are only, they only deduced one time

00:03:07.220 --> 00:03:08.840
when you look at it in factored form,

00:03:08.840 --> 00:03:12.330
only one of the factors
points to each of those zeros.

00:03:12.330 --> 00:03:15.110
So they all have a multiplicity of one.

00:03:15.110 --> 00:03:18.360
For P2, the first zero
has a multiple of one,

00:03:18.360 --> 00:03:23.020
only one of the expressions
points to a zero of one,

00:03:23.020 --> 00:03:28.020
or would become zero if
x would be equal to one.

00:03:28.040 --> 00:03:29.590
But notice, out of our factors,

00:03:29.590 --> 00:03:30.920
when we have it in factored form,

00:03:30.920 --> 00:03:33.300
out of our factored expressions,

00:03:33.300 --> 00:03:36.420
or our expression factors I should say,

00:03:36.420 --> 00:03:40.090
two of them become zero
when x is equal to three.

00:03:40.090 --> 00:03:42.290
This one and this one
are going to become zero,

00:03:42.290 --> 00:03:45.610
and so here we have a multiplicity of two.

00:03:45.610 --> 00:03:47.570
And I encourage you to
pause this video again

00:03:47.570 --> 00:03:49.300
and look at the behavior of graphs,

00:03:49.300 --> 00:03:51.360
and see if you can see a difference

00:03:51.360 --> 00:03:53.290
between the behavior of the graph

00:03:53.290 --> 00:03:55.150
when we have a multiplicity of one

00:03:55.150 --> 00:03:57.250
versus when we have a multiplicity of two.

00:03:58.470 --> 00:03:59.640
All right, now let's
look through it together.

00:03:59.640 --> 00:04:02.160
We could look at P1 where all of the zeros

00:04:02.160 --> 00:04:03.670
have a multiplicity of one,

00:04:03.670 --> 00:04:05.830
and you can see every time we have a zero

00:04:05.830 --> 00:04:07.410
we are crossing the x axis.

00:04:07.410 --> 00:04:10.260
Not only are we intersecting
it, but we are crossing it.

00:04:10.260 --> 00:04:12.870
We are crossing the x axis
there, we are crossing it again,

00:04:12.870 --> 00:04:14.570
and we're crossing it again,

00:04:14.570 --> 00:04:18.470
so at all of these we have a
sign change around that zero.

00:04:18.470 --> 00:04:19.700
But what happens here?

00:04:19.700 --> 00:04:22.270
Well on the first zero that
has a multiplicity of one,

00:04:22.270 --> 00:04:24.630
that only makes one of
the factors equal zero,

00:04:24.630 --> 00:04:27.690
we have a sign change,
just like we saw with P1.

00:04:27.690 --> 00:04:30.100
But what happens at x equals three

00:04:30.100 --> 00:04:32.880
where we have a multiplicity of two?

00:04:32.880 --> 00:04:35.720
Well there, we intersect the x axis still,

00:04:35.720 --> 00:04:40.700
P of three is zero, but notice
we don't have a sign change.

00:04:40.700 --> 00:04:43.410
We were positive before,
and we are positive after.

00:04:43.410 --> 00:04:47.930
We touch the x axis right
there, but then we go back up.

00:04:47.930 --> 00:04:50.840
And the general idea, and I
encourage you to test this out,

00:04:50.840 --> 00:04:52.850
and think about why this is true,

00:04:52.850 --> 00:04:56.600
is that if you have an odd multiplicity,

00:04:56.600 --> 00:04:57.530
now let me write this down.

00:04:57.530 --> 00:05:01.830
If the multiplicity is odd,
so if it's one, three, five,

00:05:01.830 --> 00:05:05.843
seven et cetera, then you're
going to have a sign change.

00:05:05.843 --> 00:05:07.863
Sign change.

00:05:08.730 --> 00:05:13.730
While if it is even, as the
case of two, or four, or six,

00:05:14.190 --> 00:05:16.900
you're going to have no sign change.

00:05:16.900 --> 00:05:20.060
No sign,

00:05:20.060 --> 00:05:22.183
no sign change.

00:05:23.030 --> 00:05:24.330
One way to think about it,

00:05:24.330 --> 00:05:26.610
in an example where you
have a multiplicity of two,

00:05:26.610 --> 00:05:29.830
so let's just use this zero
here, where x is equal to three,

00:05:29.830 --> 00:05:32.410
when x is less than three,

00:05:32.410 --> 00:05:34.810
both of these are going to be negative,

00:05:34.810 --> 00:05:37.060
and a negative times and
negative is a positive.

00:05:37.060 --> 00:05:39.710
And when x is greater than three,

00:05:39.710 --> 00:05:41.470
both of 'em are going to be positive,

00:05:41.470 --> 00:05:44.120
and so in either case you have a positive.

00:05:44.120 --> 00:05:47.990
So notice, you saw no sign change.

00:05:47.990 --> 00:05:50.670
Another thing to appreciate
is thinking about the number

00:05:50.670 --> 00:05:54.490
of zeros relative to the
degree of the polynomial.

00:05:54.490 --> 00:05:57.110
And what you see is is
that the number of zeros,

00:05:57.110 --> 00:05:59.730
number of zeros is

00:05:59.730 --> 00:06:02.720
at most equal to the
degree of the polynomial,

00:06:02.720 --> 00:06:05.700
so it is going to be less than or equal to

00:06:05.700 --> 00:06:10.040
the degree of the polynomial.

00:06:10.040 --> 00:06:11.800
And why is that the case?

00:06:11.800 --> 00:06:14.080
Well you might not, all
your zeros might have

00:06:14.080 --> 00:06:16.870
a multiplicity of one,
in which case the number

00:06:16.870 --> 00:06:18.810
of zeros is equal, is going to be equal

00:06:18.810 --> 00:06:21.300
to the degree of the polynomial.

00:06:21.300 --> 00:06:22.930
But if you have a zero

00:06:22.930 --> 00:06:25.250
that has a higher than one multiplicity,

00:06:25.250 --> 00:06:27.830
well then you're going to
have fewer distinct zeros.

00:06:27.830 --> 00:06:29.040
Another way to think about it is,

00:06:29.040 --> 00:06:31.603
if you were to add all the multiplicities,

00:06:31.603 --> 00:06:33.060
then that is going to be equal

00:06:33.060 --> 00:06:34.993
to the degree of your polynomial.


Zeros of polynomials: plotting zeros

WEBVTT
Kind: captions
Language: en

00:00:00.310 --> 00:00:02.200
- [Instructor] We're told
we want to find the zeros

00:00:02.200 --> 00:00:03.810
of this polynomial and they give us

00:00:03.810 --> 00:00:06.640
the polynomial right over here,
and it's in factored form.

00:00:06.640 --> 00:00:08.840
And they say plot all the zeros,

00:00:08.840 --> 00:00:11.300
or the x-intercepts, of the polynomial

00:00:11.300 --> 00:00:13.270
in the interactive graph.

00:00:13.270 --> 00:00:15.500
And so this is a screenshot
from Khan Academy.

00:00:15.500 --> 00:00:16.630
If you're doing it on Khan Academy,

00:00:16.630 --> 00:00:19.870
you would click where the
zeros are to plot the zeros,

00:00:19.870 --> 00:00:20.950
but I'm just gonna draw it in.

00:00:20.950 --> 00:00:23.350
So pause this video and
see if you could have a go

00:00:23.350 --> 00:00:26.070
at this before we work on this together.

00:00:26.070 --> 00:00:27.990
All right, now let's
work on this together.

00:00:27.990 --> 00:00:31.450
So the zeros are the x values

00:00:31.450 --> 00:00:35.450
that make our polynomial equal to zero.

00:00:35.450 --> 00:00:36.650
So another way to think about it is

00:00:36.650 --> 00:00:40.480
for what x values are
p of x equal to zero?

00:00:40.480 --> 00:00:42.730
Those would be the zeros.

00:00:42.730 --> 00:00:44.080
So essentially, we have to say,

00:00:44.080 --> 00:00:47.500
hey, what x values would
make two x times two x

00:00:47.500 --> 00:00:50.690
plus three times x minus two,

00:00:50.690 --> 00:00:52.620
'cause this is p of x, what x values would

00:00:52.620 --> 00:00:54.800
make this equal to zero?

00:00:54.800 --> 00:00:56.700
Well, as we've talked
about in previous videos,

00:00:56.700 --> 00:01:00.080
if you take the product of
things and that equals zero,

00:01:00.080 --> 00:01:02.070
if any one of those things equal zero,

00:01:02.070 --> 00:01:03.350
at least one of those things equal zero,

00:01:03.350 --> 00:01:05.370
make the whole product equal zero.

00:01:05.370 --> 00:01:08.180
So for example, if two x is equal to zero,

00:01:08.180 --> 00:01:09.820
it would make the whole thing zero,

00:01:09.820 --> 00:01:11.590
so two x could be equal to zero,

00:01:11.590 --> 00:01:13.250
and if two x is equal to zero,

00:01:13.250 --> 00:01:15.540
that means x is equal to zero,
and you could try that out.

00:01:15.540 --> 00:01:18.150
If x is equal to zero, this
part right over here is

00:01:18.150 --> 00:01:18.983
going to be equal zero.

00:01:18.983 --> 00:01:20.840
Doesn't matter what these
other two things are.

00:01:20.840 --> 00:01:22.680
Zero times something times something is

00:01:22.680 --> 00:01:24.540
going to be equal to zero.

00:01:24.540 --> 00:01:25.490
And then you could say,

00:01:25.490 --> 00:01:28.450
well, well maybe two x plus
three is equal to zero,

00:01:28.450 --> 00:01:29.410
so we could just write that.

00:01:29.410 --> 00:01:31.290
Two x plus three is equal to zero,

00:01:31.290 --> 00:01:34.100
and if that were true, what
would x, or what would x

00:01:34.100 --> 00:01:35.980
have to be in order to make that true?

00:01:35.980 --> 00:01:37.610
Subtract three from both sides,

00:01:37.610 --> 00:01:40.000
two x would have to be
equal to negative three,

00:01:40.000 --> 00:01:42.710
or x would be equal to negative 3/2.

00:01:42.710 --> 00:01:44.050
So this is another x value

00:01:44.050 --> 00:01:45.460
that would make the whole thing zero,

00:01:45.460 --> 00:01:49.370
'cause if x is equal to negative
3/2, then two x plus three

00:01:49.370 --> 00:01:52.440
is equal to zero, you take a
zero times whatever this is

00:01:52.440 --> 00:01:54.600
and whatever that is,
you're gonna get zero.

00:01:54.600 --> 00:01:56.770
And then last but not least,

00:01:56.770 --> 00:01:58.420
x minus two could be equal to zero.

00:01:58.420 --> 00:02:01.070
That would make the whole
product equal to zero.

00:02:01.070 --> 00:02:04.360
So what x value makes
x minus two equal zero?

00:02:04.360 --> 00:02:05.740
We'll add two to both sides,

00:02:05.740 --> 00:02:07.700
and you would get x is equal to two.

00:02:07.700 --> 00:02:09.910
If x equals two, that equals zero,

00:02:09.910 --> 00:02:11.850
doesn't matter what these
other two things are.

00:02:11.850 --> 00:02:13.930
Zero times something times something is

00:02:13.930 --> 00:02:15.670
going to be equal to zero.

00:02:15.670 --> 00:02:19.560
So just like that, we have
the zeros of our polynomial,

00:02:19.560 --> 00:02:20.530
and the reason why they have

00:02:20.530 --> 00:02:22.880
x-intercepts in parentheses here

00:02:22.880 --> 00:02:25.410
is that's where the graph of p of x,

00:02:25.410 --> 00:02:27.010
if you say y equals p of x,

00:02:27.010 --> 00:02:29.090
that's where it would
intersect the x-axis,

00:02:29.090 --> 00:02:30.160
and that's because that's

00:02:30.160 --> 00:02:33.120
where our polynomial is equal to zero.

00:02:33.120 --> 00:02:35.670
So let's see, we have x equal zero

00:02:35.670 --> 00:02:36.940
which is right over there.

00:02:36.940 --> 00:02:38.530
Once again, if you're
doing this on Khan Academy,

00:02:38.530 --> 00:02:39.680
you would just click right over there

00:02:39.680 --> 00:02:41.410
and it would put a little dot there.

00:02:41.410 --> 00:02:43.820
We have x is equal to negative 3/2,

00:02:43.820 --> 00:02:45.670
which is the same thing as negative 1/2,

00:02:45.670 --> 00:02:47.690
so that's right over there.

00:02:47.690 --> 00:02:52.690
And then, we have x equals
two, which is right over there.

00:02:53.000 --> 00:02:54.330
So those are the x-intercepts

00:02:54.330 --> 00:02:57.340
or the zeros of that polynomial.

00:02:57.340 --> 00:02:58.930
Now, this is useful in life,

00:02:58.930 --> 00:03:02.090
because you could use
it to graph a function.

00:03:02.090 --> 00:03:03.880
I don't know exactly what
this function looks like,

00:03:03.880 --> 00:03:06.810
maybe it looks something like this,

00:03:06.810 --> 00:03:10.470
maybe it looks something like this.

00:03:10.470 --> 00:03:12.700
We would have to try out a few
other values to get a sense

00:03:12.700 --> 00:03:14.620
of that, but we at least know

00:03:14.620 --> 00:03:16.560
where it's intersecting the x-axis.

00:03:16.560 --> 00:03:17.913
It's at the zeros.


Zeros of polynomials (with factoring): grouping

WEBVTT
Kind: captions
Language: en

00:00:00.360 --> 00:00:01.540
- [Instructor] So we're told that p of x

00:00:01.540 --> 00:00:03.010
is equal to this expression here,

00:00:03.010 --> 00:00:06.300
and it says plot all the
zeros or x-intercepts,

00:00:06.300 --> 00:00:08.750
of the polynomial in
the interactive graph,

00:00:08.750 --> 00:00:10.760
and the reason why it
says interactive graph

00:00:10.760 --> 00:00:13.010
is this is a screenshot from this type

00:00:13.010 --> 00:00:14.230
of exercise on Khan Academy,

00:00:14.230 --> 00:00:16.780
and on Khan Academy, you'd be
able to click on points here

00:00:16.780 --> 00:00:17.760
and it'd put little dots

00:00:17.760 --> 00:00:20.210
and you can either delete them
and put 'em someplace else,

00:00:20.210 --> 00:00:21.370
so it'd be an interactive graph.

00:00:21.370 --> 00:00:22.480
But this is just a screenshot,

00:00:22.480 --> 00:00:24.440
so I'm just going to draw on top of this.

00:00:24.440 --> 00:00:27.240
But the main goal is, what are
the zeros of this polynomial

00:00:27.240 --> 00:00:29.300
and then you just to
plot it on this graph,

00:00:29.300 --> 00:00:31.300
so pause this video and have a go at it.

00:00:32.240 --> 00:00:35.760
All right, now to figure out
the zeros of a polynomial,

00:00:35.760 --> 00:00:37.270
you would essentially have
to figure out the x values

00:00:37.270 --> 00:00:39.620
that would make the
polynomial equal to zero.

00:00:39.620 --> 00:00:41.860
Or another way to think
about it is the x values

00:00:41.860 --> 00:00:43.520
that would make this equation true.

00:00:43.520 --> 00:00:47.090
X to the third plus x squared

00:00:47.980 --> 00:00:52.470
minus nine x minus nine is equal to zero.

00:00:52.470 --> 00:00:53.570
Now, the best way to do that

00:00:53.570 --> 00:00:56.830
is to try to factor this expression.

00:00:56.830 --> 00:00:58.760
Now, this is a third degree polynomial,

00:00:58.760 --> 00:01:00.920
which isn't always so easy to factor,

00:01:00.920 --> 00:01:03.000
so let's see how we might approach it.

00:01:03.000 --> 00:01:05.430
The first thing I look for is
are there any common factors

00:01:05.430 --> 00:01:09.360
to all of these terms, and it
doesn't look like there is.

00:01:09.360 --> 00:01:12.480
The next thing I could look for,

00:01:12.480 --> 00:01:13.930
I could think about whether factoring

00:01:13.930 --> 00:01:16.180
by grouping could work here,

00:01:16.180 --> 00:01:17.700
and when I think about
factoring by grouping,

00:01:17.700 --> 00:01:20.940
I would look at the first
two terms and I would

00:01:20.940 --> 00:01:22.937
look at the last two terms,

00:01:22.937 --> 00:01:25.130
and I would say, is there
anything I could factor out

00:01:25.130 --> 00:01:28.500
of these first two terms that would,

00:01:28.500 --> 00:01:30.090
or what's the most that I could factor out

00:01:30.090 --> 00:01:31.170
of these first two terms,

00:01:31.170 --> 00:01:32.410
and then what's the most
that I could factor out

00:01:32.410 --> 00:01:33.480
of these last two terms,

00:01:33.480 --> 00:01:34.410
and then it would leave

00:01:34.410 --> 00:01:37.000
something similar once
I've done that factoring.

00:01:37.000 --> 00:01:39.910
Now, what I mean is, for these first two,

00:01:39.910 --> 00:01:41.850
we have a common factor of x squared,

00:01:41.850 --> 00:01:43.530
so let's factor out an x squared

00:01:43.530 --> 00:01:48.290
and these first two terms become
x squared times x plus one,

00:01:48.290 --> 00:01:51.180
and then for these second two terms,

00:01:51.180 --> 00:01:53.510
I can factor out a negative nine,

00:01:53.510 --> 00:01:58.510
so I could rewrite it as
negative nine times x plus one.

00:01:58.910 --> 00:02:01.470
Now, that all worked out quite nicely,

00:02:01.470 --> 00:02:04.660
because now we see, if we view,

00:02:04.660 --> 00:02:07.380
if we view this as our now our first term

00:02:07.380 --> 00:02:09.000
and this as our second term,

00:02:09.000 --> 00:02:14.000
we can see that x plus one
is a factor of both of them.

00:02:14.350 --> 00:02:16.950
And so we can factor that out.

00:02:16.950 --> 00:02:19.900
We can factor out the x plus one,

00:02:19.900 --> 00:02:22.040
and I'll do that in this light blue color,

00:02:22.040 --> 00:02:24.730
actually let me do it with
slightly darker blue color.

00:02:24.730 --> 00:02:26.810
And so if you factor out the x plus one,

00:02:26.810 --> 00:02:31.810
you're left with x plus
one times x squared,

00:02:32.820 --> 00:02:36.460
x squared, minus nine.

00:02:36.460 --> 00:02:38.100
Minus nine.

00:02:38.100 --> 00:02:41.710
And that is going to be equal to zero.

00:02:41.710 --> 00:02:43.530
Now, we are not done factoring yet,

00:02:43.530 --> 00:02:45.770
because now we have a
difference of squares.

00:02:45.770 --> 00:02:48.510
X squared minus nine, this
is going to be equal to,

00:02:48.510 --> 00:02:50.010
and let me just write it all out,

00:02:50.010 --> 00:02:52.320
so I have this x plus one here,

00:02:52.320 --> 00:02:57.320
so I have x plus one, and
then the x squared minus nine,

00:02:57.610 --> 00:03:02.610
I can write as x plus
three times x minus three.

00:03:02.870 --> 00:03:05.930
If any of what I'm doing
feels unfamiliar to you,

00:03:05.930 --> 00:03:08.490
if that first factoring feels unfamiliar,

00:03:08.490 --> 00:03:10.880
I encourage you to review
factoring by grouping,

00:03:10.880 --> 00:03:13.010
and if what I just did looks unfamiliar,

00:03:13.010 --> 00:03:16.230
I encourage you to look at
factoring differences of squares.

00:03:16.230 --> 00:03:19.840
But anyway, all of that
would be equal to zero.

00:03:19.840 --> 00:03:23.930
Now, if I have the product of
several things equaling zero,

00:03:23.930 --> 00:03:26.560
any, if any one of those
things is equal to zero,

00:03:26.560 --> 00:03:28.840
that would make the whole
expression equal to zero.

00:03:28.840 --> 00:03:31.320
So we have a situation where one

00:03:31.320 --> 00:03:32.380
solution would be the solution

00:03:32.380 --> 00:03:35.350
that makes x plus one equal to zero,

00:03:35.350 --> 00:03:37.290
and once again, I'm gonna
do that in darker color,

00:03:37.290 --> 00:03:39.700
x plus one equal zero, and

00:03:39.700 --> 00:03:42.380
that of course is x is
equal to negative one.

00:03:42.380 --> 00:03:44.470
Another solution is what would make x

00:03:44.470 --> 00:03:47.220
plus three equal to zero?

00:03:47.220 --> 00:03:49.550
And that of course is x is
equal to negative three,

00:03:49.550 --> 00:03:51.160
subtract three from both sides,

00:03:51.160 --> 00:03:53.770
and then another solution is

00:03:53.770 --> 00:03:58.680
going to be whatever x value
makes x minus three equal

00:03:58.680 --> 00:04:00.250
to zero, add three to both sides,

00:04:00.250 --> 00:04:02.600
you get x is equal to three.

00:04:02.600 --> 00:04:03.630
So there you have it.

00:04:03.630 --> 00:04:05.450
We have our three zeros.

00:04:05.450 --> 00:04:06.980
Our polynomial evaluated

00:04:06.980 --> 00:04:10.330
at any of these x values
will be equal to zero.

00:04:10.330 --> 00:04:13.740
So we can plot it here on
this interactive graph,

00:04:13.740 --> 00:04:15.150
I'm just gonna draw on it.

00:04:15.150 --> 00:04:16.840
So we have x equals negative one,

00:04:16.840 --> 00:04:18.860
which is right over there,

00:04:18.860 --> 00:04:20.980
x equals negative three,

00:04:20.980 --> 00:04:22.700
which is right over there,

00:04:22.700 --> 00:04:25.780
and x equals three, which
is right over there.

00:04:25.780 --> 00:04:28.810
And the reason why you might
want to do this type of thing,

00:04:28.810 --> 00:04:30.660
this exercise just asks us to do this,

00:04:30.660 --> 00:04:32.270
and we're done, but the reason why this

00:04:32.270 --> 00:04:35.140
is useful is this can help
inform what the graph looks like.

00:04:35.140 --> 00:04:38.090
This tells us where our
graph intersects the x-axis.

00:04:38.090 --> 00:04:41.503
So our graph might do something like this,

00:04:42.600 --> 00:04:47.600
or it might do something like this,

00:04:47.600 --> 00:04:48.800
and we would have to look

00:04:48.800 --> 00:04:52.470
at other information to think
about what that might be.

00:04:52.470 --> 00:04:53.793
But I'll leave you there.


Zeros of polynomials (with factoring): common factor

WEBVTT
Kind: captions
Language: en

00:00:00.530 --> 00:00:04.020
- So we're given a p of x,
it's a third degree polynomial,

00:00:04.020 --> 00:00:07.120
and they say, plot all the
zeroes or the x-intercepts

00:00:07.120 --> 00:00:09.310
of the polynomial in
the interactive graph.

00:00:09.310 --> 00:00:10.510
And the reason why they
say interactive graph,

00:00:10.510 --> 00:00:13.810
this is a screen shot from
the exercise on Kahn Academy,

00:00:13.810 --> 00:00:16.810
where you could click
and place the zeroes.

00:00:16.810 --> 00:00:20.340
But the key here is, lets
figure out what x values make

00:00:20.340 --> 00:00:22.640
p of x equal to zero,

00:00:22.640 --> 00:00:23.960
those are the zeroes.

00:00:23.960 --> 00:00:25.230
And then we can plot them.

00:00:25.230 --> 00:00:26.063
So pause this video,

00:00:26.063 --> 00:00:28.380
and see if you can figure that out.

00:00:28.380 --> 00:00:32.660
So the key here is to try
to factor this expression

00:00:32.660 --> 00:00:34.320
right over here, this
third degree expression,

00:00:34.320 --> 00:00:36.850
because really we're
trying to solve the X's

00:00:36.850 --> 00:00:40.490
for which five x to
third plus five x squared

00:00:40.490 --> 00:00:43.070
minus 30 x is equal to zero.

00:00:43.070 --> 00:00:44.640
And the way we do that is

00:00:44.640 --> 00:00:47.210
by factoring this left-hand expression.

00:00:47.210 --> 00:00:48.890
So the first thing I always look for

00:00:48.890 --> 00:00:51.140
is a common factor
across all of the terms.

00:00:51.140 --> 00:00:55.240
It looks like all of the
terms are divisible by five x.

00:00:55.240 --> 00:00:57.170
So let's factor out a five x.

00:00:57.170 --> 00:00:58.980
So this is going to be five x times,

00:00:58.980 --> 00:01:01.370
if we take a five x out
of five x to the third,

00:01:01.370 --> 00:01:03.040
we're left with an x squared.

00:01:03.040 --> 00:01:04.840
If we take out a five x
out of five x squared,

00:01:04.840 --> 00:01:06.800
we're left with an x, so plus x.

00:01:06.800 --> 00:01:09.200
And if we take out a
five x of negative 30 x,

00:01:09.200 --> 00:01:13.230
we're left with a negative
six is equal to zero.

00:01:13.230 --> 00:01:17.800
And now, we have five x
times this second degree,

00:01:17.800 --> 00:01:20.580
the second degree expression
and to factor that,

00:01:20.580 --> 00:01:23.230
let's see, what two numbers add up to one?

00:01:23.230 --> 00:01:25.110
You could use as a one x here.

00:01:25.110 --> 00:01:27.730
And their product is
equal to negative six.

00:01:27.730 --> 00:01:31.290
And let's see, positive
three and negative two

00:01:31.290 --> 00:01:32.550
would do the trick.

00:01:32.550 --> 00:01:35.400
So I can rewrite this as five x times,

00:01:35.400 --> 00:01:39.510
so x plus three, x plus three,

00:01:39.510 --> 00:01:43.410
times x minus two, and if
what I did looks unfamiliar,

00:01:43.410 --> 00:01:46.100
I encourage you to review
factoring quadratics

00:01:46.100 --> 00:01:47.320
on Kahn Academy,

00:01:47.320 --> 00:01:50.460
and that is all going to be equal to zero.

00:01:50.460 --> 00:01:53.170
And so if I try to
figure out what x values

00:01:53.170 --> 00:01:54.780
are going to make this
whole expression zero,

00:01:54.780 --> 00:01:56.185
it could be the x values

00:01:56.185 --> 00:01:58.910
or the x value that
makes five x equal zero.

00:01:58.910 --> 00:02:00.210
Because if five x zero,

00:02:00.210 --> 00:02:02.980
zero times anything else
is going to be zero.

00:02:02.980 --> 00:02:06.000
So what makes five x equal zero?

00:02:06.000 --> 00:02:09.550
Well if we divide five, if
you divide both sides by five,

00:02:09.550 --> 00:02:11.340
you're going to get x is equal to zero.

00:02:11.340 --> 00:02:12.450
And it is the case.

00:02:12.450 --> 00:02:15.230
If x equals zero, this becomes zero,

00:02:15.230 --> 00:02:17.240
and then doesn't matter what these are,

00:02:17.240 --> 00:02:18.976
zero times anything is zero.

00:02:18.976 --> 00:02:22.050
The other possible x value
that would make everything zero

00:02:22.050 --> 00:02:25.460
is the x value that makes
x plus three equal to zero.

00:02:25.460 --> 00:02:26.920
Subtract three from both sides

00:02:26.920 --> 00:02:29.000
you get x is equal to negative three.

00:02:29.000 --> 00:02:31.420
And then the other x value
is the x value that makes

00:02:31.420 --> 00:02:34.050
x minus two equal to zero.

00:02:34.050 --> 00:02:37.210
Add two to both sides,
that's gonna be x equals two.

00:02:37.210 --> 00:02:38.043
So there you have it.

00:02:38.043 --> 00:02:41.150
We have identified three x
values that make our polynomial

00:02:41.150 --> 00:02:43.250
equal to zero and those
are going to be the zeros

00:02:43.250 --> 00:02:44.770
and the x intercepts.

00:02:44.770 --> 00:02:47.830
So we have one at x equals zero.

00:02:47.830 --> 00:02:52.260
We have one at x equals negative three.

00:02:52.260 --> 00:02:57.260
We have one at x equals, at x equals two.

00:02:58.620 --> 00:02:59.830
And the reason why it's,

00:02:59.830 --> 00:03:01.370
we're done now with this exercise,

00:03:01.370 --> 00:03:02.580
if you're doing this on Kahn Academy

00:03:02.580 --> 00:03:04.460
or just clicked in these three places,

00:03:04.460 --> 00:03:07.201
but the reason why folks
find this to be useful is

00:03:07.201 --> 00:03:10.320
it helps us start to think
about what the graph could be.

00:03:10.320 --> 00:03:12.140
Because the graph has to intercept

00:03:12.140 --> 00:03:13.904
the x axis at these points.

00:03:13.904 --> 00:03:18.360
So the graph might look
something like that,

00:03:18.360 --> 00:03:21.680
it might look something like that.

00:03:21.680 --> 00:03:24.020
And to figure out what it
actually does look like

00:03:24.020 --> 00:03:26.330
we'd probably want to try
out a few more x values

00:03:26.330 --> 00:03:28.300
in between these x intercepts

00:03:28.300 --> 00:03:31.083
to get the general sense of the graph.