Is $x + 1$ $x+1$ a factor of $x^{3} + 8 x^{2} + 11 x - 20$ $x^3+8x^2+11x-20$ ?

Question

2803 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-08T13:02:29

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

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

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

$p (- 1) = 0$ $p(-1)=0$

Verifying on $p (x)$ $p(x)$

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

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

but $(x - 1)$ $(x-1)$ is a factor because

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

Is x+1x+1x+1 a factor of x^3+8x^2+11x-20x3+8x2+11x−20x^3+8x^2+11x-20?