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

May 24, 2017

$f \left(x\right) = \left(x + 3\right) \left({x}^{2} - 2 x + 8\right) .$

#### Explanation:

We have, for, $f \left(x\right) = {x}^{3} + {x}^{2} + 2 x + 24 ,$

$f \left(- 3\right) = - 27 + 9 - 6 + 24 = 0.$

From the Factor Theorem, hence, we conclude that,

$x - \left(- 3\right) = x + 3$ is a factor of $f \left(x\right) .$

One way, then, is to divide $f \left(x\right)$ by $\left(x + 3\right)$ by Long Division.

Instead, have a look at the following :

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

$= \underline{{x}^{3} + 3 {x}^{2}} - \underline{2 {x}^{2} - 6 x} + \underline{8 x + 24} ,$

$= {x}^{2} \left(x + 3\right) - 2 x \left(x + 3\right) + 8 \left(x + 3\right) ,$

$\Rightarrow f \left(x\right) = \left(x + 3\right) \left({x}^{2} - 2 x + 8\right) .$

For the Quadr. Poly. ${x}^{2} - 2 x + 8 ,$ we have,

$\Delta = {b}^{2} - 4 a c = 4 - 4 \left(1\right) \left(8\right) = - 28 < 0.$

Hence, it can not be factored further in $\mathbb{R} .$

Enjoy Maths.!