Question

Solve: $x^3-3x^2-9x-5$

Answer 1

Let $f\left(x\right)=x^3-3x^2-9x-5$ be the given polynomial.

Now, putting $x=-1$, we get

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

$=-1-3+9-5=-9+9=0$

Therefore, $(x+1)$ is a factor of polynomial f(x).

Now,

$f\left(x\right)=x^3-3x^2-9x-5$

$=x^3+x^2-4x^2-4x-5x-5$

$=x^2(x+1)-4x(x+1)-5(x+1)$

$=(x+1)(x^2-4x-5)$

$=(x+1)(x^2-5x+x-5)$

$=(x+1)\left\{x\right(x-5)+1(x-5\left)\right\}$

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

Hence $(x+1),(x+1),(x-5)$ are the factors of polynomial f(x).