# What is the remainder when the function f(x)=x^3-4x^2+12 is divided by (x+2)?

Jan 11, 2018

$\textcolor{b l u e}{- 12}$

#### Explanation:

The Remainder theorem states that, when $f \left(x\right)$ is divided by $\left(x - a\right)$

$f \left(x\right) = g \left(x\right) \left(x - a\right) + r$

Where $g \left(x\right)$ is the quotient and $r$ is the remainder.

If for some $x$ we can make $g \left(x\right) \left(x - a\right) = 0$, then we have:

$f \left(a\right) = r$

From example:

${x}^{3} - 4 {x}^{2} + 12 = g \left(x\right) \left(x + 2\right) + r$

Let $x = - 2$

$\therefore$

${\left(- 2\right)}^{3} - 4 {\left(- 2\right)}^{2} + 12 = g \left(x\right) \left(\left(- 2\right) + 2\right) + r$

$- 12 = 0 + r$

$\textcolor{b l u e}{r = - 12}$

This theorem is just based on what we know about numerical division. i.e.

The divisor x the quotient + the remainder = the dividend

$\therefore$

$\frac{6}{4} = 1$ + remainder 2.

$4 \times 1 + 2 = 6$

Jan 11, 2018

$\text{remainder } = - 12$

#### Explanation:

$\text{using the "color(blue)"remainder theorem}$

$\text{the remainder when "f(x)" is divided by "(x-a)" is } f \left(a\right)$

$\text{here } \left(x - a\right) = \left(x - \left(- 2\right)\right) \Rightarrow a = - 2$

$f \left(- 2\right) = {\left(- 2\right)}^{3} - 4 {\left(- 2\right)}^{2} + 12 = - 12$