# How do you use the remainder theorem to find the remainder for the division (n^4-n^3-10n^2+4n+24)div(n+2)?

Oct 10, 2016

f(-2) = 0""
There is no remainder and $\left(n + 2\right)$ is a factor of ${n}^{4} - {n}^{3} - 10 {n}^{2} + 4 n + 24$

#### Explanation:

Let $f \left(n\right) = {n}^{4} - {n}^{3} - 10 {n}^{2} + 4 n + 24$

Setting $n + 2 = 0 \text{ }$ gives $n = - 2$

Substitute $n = - 2 \text{ into " f (n)}$ to find the remainder.

$f \left(- 2\right) = {\left(- 2\right)}^{4} - {\left(- 2\right)}^{3} - 10 {\left(- 2\right)}^{2} + 4 \left(- 2\right) + 24$

$f \left(- 2\right) = 16 + 8 - 40 - 8 + 24 = 0$

As this is 0, it means there is no remainder.

This tells us that $n + 2$ is a factor of ${n}^{4} - {n}^{3} - 10 {n}^{2} + 4 n + 24$ and therefore divides into it exactly without leaving a remainder.