# Is x+1 a factor of x^3+8x^2+11x-20?

Nov 8, 2016

$\left(x + 1\right)$ is not a factor, but $\left(x - 1\right)$ is.

#### Explanation:

Given $p \left(x\right) = {x}^{3} + 8 {x}^{2} + 11 x - 20$ if $x + 1$ is a factor of $p \left(x\right)$ then

$p \left(x\right) = \left(x + 1\right) q \left(x\right)$ so for $x = - 1$ we must have

$p \left(- 1\right) = 0$

Verifying on $p \left(x\right)$

$p \left(- 1\right) = {\left(- 1\right)}^{3} + 8 {\left(- 1\right)}^{2} + 11 \left(- 1\right) - 20 = - 24$

so $\left(x + 1\right)$ is not a factor of $p \left(x\right)$

but $\left(x - 1\right)$ is a factor because

$p \left(1\right) = 1 + 8 + 11 - 20 = 0$