# How do you use the remainder theorem to evaluate f(a)=a^3+3a^2+2a+8 at a=-3?

Dec 3, 2016

$f \left(- 3\right) = 2$

#### Explanation:

$f \left(a\right) = {a}^{3} + 3 {a}^{2} + 2 a + 8$ at $a = - 3$

$\underline{- 3} | 1 \textcolor{w h i t e}{a a a a a} 3 \textcolor{w h i t e}{a a a} 2 \textcolor{w h i t e}{a a a a} 8$
color(white)(aaaa)ul{darrcolor(white)(aa)-3color(white)(aaa)0color(white)(a)-6
$\textcolor{w h i t e}{{a}^{2} a a a} 1 \textcolor{w h i t e}{a a a a a} 0 \textcolor{w h i t e}{a a a} 2 \textcolor{w h i t e}{a {a}^{2} a} \textcolor{red}{2}$

The last number $\textcolor{red}{2}$ is the remainder and is also the
value of $f \left(- 3\right) .$

$f \left(- 3\right) = \textcolor{red}{2}$

You can check your solution by substituting $x = - 3$.

$f \left(- 3\right) = {\left(- 3\right)}^{3} + 3 {\left(- 3\right)}^{2} + 2 \left(- 3\right) + 8 = 2$