# How do you use the remainder theorem to evaluate f(x)=x^5-47x^3-16x^2+8x+52 at x=7?

Jan 16, 2017

$f \left(7\right) = 10$

#### Explanation:

Remainder theorem states that if a polynomial $f \left(x\right)$ is divided by $\left(x - a\right)$, the the remainder is $f \left(a\right)$.

Hence to evaluate $f \left(x\right) = {x}^{5} - 47 {x}^{3} - 16 {x}^{2} + 8 x + 52$ at $x = 7$, we need to divide $f \left(x\right) = {x}^{5} - 47 {x}^{3} - 16 {x}^{2} + 8 x + 52$ by $\left(x - 7\right)$.

Now, $f \left(x\right) = {x}^{5} - 47 {x}^{3} - 16 {x}^{2} + 8 x + 52$

$= {x}^{4} \left(x - 7\right) + 7 {x}^{3} \left(x - 7\right) + 2 {x}^{2} \left(x - 7\right) - 2 x \left(x - 7\right) - 6 \left(x - 7\right) + 10$

$= \left({x}^{4} + 7 {x}^{3} + 2 {x}^{2} - 2 x - 6\right) \left(x - 7\right) + 10$

As such the remainder on dividing $f \left(x\right)$ by $\left(x - 7\right)$ is $10$

hence $f \left(7\right) = 10$

Check $f \left(6\right) = {7}^{5} - 47 \times {7}^{3} - 16 \times {7}^{2} + 8 \times 7 + 52$

$= 16807 - 16121 - 784 + 56 + 52 = 10$