# How do you factor given that f(-14)=0 and f(x)=x^3+11x^2-150x-1512?

Oct 5, 2016

$\left(x - 12\right) \left(x + 9\right) \left(x + 14\right)$

#### Explanation:

When you know that $f \left(c\right) = 0$ for some value $c$, you knot that $f \left(x\right)$ has a factor $\left(x - c\right)$. So, you can divide ${x}^{3} + 11 {x}^{2} - 150 x - 1512$ by $\left(x + 14\right)$ via long division, or whatever method you prefer, and obtain

$\frac{{x}^{3} + 11 {x}^{2} - 150 x - 1512}{x + 14} = {x}^{2} - 3 x - 108$

which means

$\left({x}^{2} - 3 x - 108\right) \left(x + 14\right) = {x}^{3} + 11 {x}^{2} - 150 x - 1512$

Again, ${x}^{2} - 3 x - 108$ has solutions $12$ and $- 9$, which means that it can be written as $\left(x - 12\right) \left(x + 9\right)$

Thus, we can write

${x}^{3} + 11 {x}^{2} - 150 x - 1512 = \left({x}^{2} - 3 x - 108\right) \left(x + 14\right) = \left(x - 12\right) \left(x + 9\right) \left(x + 14\right)$