# How do you use the remainder theorem to evaluate f(a)=a^4+3a^3-17a^2+2a-7 at a=3?

Jan 31, 2017

$f \left(a\right) = {a}^{4} + 3 {a}^{3} - 17 {a}^{2} + 2 a - 7$ at $a = 3$ is $8$

#### Explanation:

According to remainder theorem if a polynomial $f \left(x\right)$ is divided by $\left(x - p\right)$, the remainder is $f \left(p\right)$.

Hence, using remainder theorem to evaluate $f \left(a\right) = {a}^{4} + 3 {a}^{3} - 17 {a}^{2} + 2 a - 7$ at $a = 3$,

we should divide ${a}^{4} + 3 {a}^{3} - 17 {a}^{2} + 2 a - 7$ by $\left(a - 3\right)$

this can be done using synthetic division

$3 | \textcolor{w h i t e}{X} 1 \text{ "color(white)(X)3color(white)(XX)-17" "" } 2 \textcolor{w h i t e}{X X} - 7$
$\textcolor{w h i t e}{x} | \text{ } \textcolor{w h i t e}{X \times} 3 \textcolor{w h i t e}{X X X} 18 \textcolor{w h i t e}{X \times x} 3 \textcolor{w h i t e}{X \times x} 15$
" "stackrel("—————————————----)
$\textcolor{w h i t e}{x} | \textcolor{w h i t e}{X} \textcolor{b l u e}{1} \textcolor{w h i t e}{X 11} \textcolor{red}{6} \textcolor{w h i t e}{X X X X} \textcolor{red}{1} \textcolor{w h i t e}{X X X} \textcolor{red}{5} \textcolor{w h i t e}{X \times x} 8$

Hence, Quotient is ${a}^{3} + 6 {a}^{2} + a + 5$ and remainder is $8$.

Hence $f \left(a\right) = {a}^{4} + 3 {a}^{3} - 17 {a}^{2} + 2 a - 7$ at $a = 3$ is $8$