Question

Factorize the given polynomial
$x^3-3x^2-9x-5$

Answer 1

Given $x^3-3x^2-9x-5$

Let $x=1$

So, $x^3-3x^2-9x-5$

$=(1{)}^3-3(1{)}^2-9\left(1\right)-5$

$=1-3-9-5$

$=-16\ne 0$

So, $1$ is not a factor

Let $x=-1$

So, $x^3-3x^2-9x-5$

$=(-1{)}^3-3(-1{)}^2-9(-1)-5$

$=-1-3+9-5$

$=0$

So, $-1$ is a factor

Divide $x^3-3x^2-9x-5$ by $(x+1)$

(Refer image)

$\therefore x^2-4x-5=x^2-5x+x-5$

$=x(x-5)+1(x-5)$

$=(x-5)(x+1)$

So, factors are $(x+1)(x-5)(x+1)$.