# How do you factor given that f(6)=0 and f(x)=x^3-3x^2-16x-12?

Sep 4, 2017

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

#### Explanation:

$\text{using the "color(blue)"factor theorem}$

•color(white)(x)f(a)=0" if and only if "(x-a)" is a factor of "f(x)

$f \left(6\right) = 0 \Rightarrow \left(x - 6\right) \text{ is a factor of } f \left(x\right)$

$f \left(x\right) = \textcolor{red}{{x}^{2}} \left(x - 6\right) \textcolor{m a \ge n t a}{+ 6 {x}^{2}} - 3 {x}^{2} - 16 x - 12$

$\textcolor{w h i t e}{f \left(x\right)} = \textcolor{red}{{x}^{2}} \left(x - 6\right) \textcolor{red}{+ 3 x} \left(x - 6\right) \textcolor{m a \ge n t a}{+ 18 x} - 16 x - 12$

$\textcolor{w h i t e}{f \left(x\right)} = \textcolor{red}{{x}^{2}} \left(x - 6\right) \textcolor{red}{+ 3 x} \left(x - 6\right) \textcolor{red}{+ 2} \left(x - 6\right) \textcolor{m a \ge n t a}{+ 12} - 12$

$\textcolor{w h i t e}{f \left(x\right)} = \textcolor{red}{{x}^{2}} \left(x - 6\right) \textcolor{red}{+ 3 x} \left(x - 6\right) \textcolor{red}{+ 2} \left(x - 6\right) + 0$

$\Rightarrow f \left(x\right) = \left(x - 6\right) \left(\textcolor{red}{{x}^{2} + 3 x + 2}\right)$

$\textcolor{w h i t e}{\Rightarrow f \left(x\right)} = \left(x - 6\right) \left(x + 1\right) \left(x + 2\right)$