# What are all the possible rational zeros for f(x)=x^3+11x^2+35x+33 and how do you find all zeros?

##### 1 Answer
Dec 26, 2016

$- 3$; (rational zero)
$- 4 \pm \sqrt{5}$ (not rational zeros)

#### Explanation:

You would find rational zeros in the set of integer and negative numbers dividing the known term 33, which are -1; -3; -11; -33.

Using the remainder theorem, you will find that

$f \left(- 1\right) \ne 0$

but

$f \left(- 3\right) = 0$

So -3 is a rational zero.

Then you will divide:

$\left({x}^{3} + 11 {x}^{2} + 35 x + 33\right) : \left(x + 3\right) = {x}^{2} + 8 x + 11$

By using quadratic formula, you would find the not rational zeros:

$- 4 \pm \sqrt{5}$