# How do you factor given that f(-2)=0 and f(x)=x^3-5x^2-2x+24?

Sep 20, 2016

$f \left(x\right) = \left(x + 2\right) \left(x - 4\right) \left(x - 3\right)$.

#### Explanation:

Having verified $f \left(- 2\right) = 0 , \left(x + 2\right)$ is a factor of $f \left(x\right) .$

Long Division is well-known Method to factorise, but we proceed as under :-

$f \left(x\right) = {x}^{3} - 5 {x}^{2} - 2 x + 24$

$= \underline{{x}^{3} + 2 {x}^{2}} - \underline{7 {x}^{2} - 14 x} + \underline{12 x + 24}$

$= {x}^{2} \left(x + 2\right) - 7 x \left(x + 2\right) + 12 \left(x + 2\right)$

$= \left(x + 2\right) \left({x}^{2} - 7 x + 12\right)$

$= \left(x + 2\right) \left\{\underline{{x}^{2} - 4 x} - \underline{3 x + 12}\right\}$

$= \left(x + 2\right) \left\{x \left(x - 4\right) - 3 \left(x - 4\right)\right\}$

$= \left(x + 2\right) \left(x - 4\right) \left(x - 3\right)$.

Enjoy Maths.!