# How do you factor given that f(-6)=0 and f(x)=2x^3+7x^2-33x-18?

Aug 8, 2018

$f \left(x\right) = \left(x + 6\right) \left(x - 3\right) \left(2 x + 1\right)$.

#### Explanation:

The given cubic polynomial $f$ has $- 6$ as its zero.

$\therefore \left(x - \left(- 6\right)\right) = \left(x + 6\right)$ is a factor of $f \left(x\right)$.

We split the terms of $f \left(x\right)$ in such a way that $\left(x + 6\right)$ may turn

out as a factor, as shown below :-

$f \left(x\right) = 2 {x}^{3} + 7 {x}^{2} - 33 x - 18$,

$= \underline{2 {x}^{3} + 12 {x}^{2}} - \underline{5 {x}^{2} - 30 x} - \underline{3 x - 18}$,

$= 2 {x}^{2} \left(x + 6\right) - 5 x \left(x + 6\right) - 3 \left(x + 6\right)$,

$= \left(x + 6\right) \left\{2 {x}^{2} - 5 x - 3\right\}$,

$= \left(x + 6\right) \left\{\underline{2 {x}^{2} - 6 x} + \underline{x - 3}\right\}$,

$= \left(x + 6\right) \left\{2 x \left(x - 3\right) + 1 \left(x - 3\right)\right\}$.

$\Rightarrow f \left(x\right) = \left(x + 6\right) \left(x - 3\right) \left(2 x + 1\right)$.