# How do you use the remainder theorem to see if the p+5 is a factor of p^4+6p^3+11p^2+29p-13?

Feb 8, 2017

Not a factor.

#### Explanation:

Before even using the remainder theorem, you can see that $\left(p + 5\right)$ will not be a factor because the expression ends with $- 13$ which is not divisible by 5.

But looking at it again using the remainder theorem...

Call The expression $f \left(p\right)$

$f \left(p\right) = {p}^{4} + 6 {p}^{3} + 11 {p}^{2} + 29 p - 13$

If $p + 5 = 0 \rightarrow p = - 5$

Calculate $f \left(- 5\right)$ by substituting (-5) for every $p$

If the answer is equal to $0$, it means that $\left(p + 5\right)$ is a factor.

If the answer is not $0$, then the value you get will be the remainder if you divide by $\left(p + 5\right)$

$f \left(p\right) = {p}^{4} + 6 {p}^{3} + 11 {p}^{2} + 29 p - 13$

$f \left(- 5\right) = {\left(- 5\right)}^{4} + 6 {\left(- 5\right)}^{3} + 11 {\left(- 5\right)}^{2} + 29 \left(- 5\right) - 13$

$= 625 - 750 + 275 - 145 - 13$

$= - 8$

Nope, not a factor!