# How do you find all zeros of the function f(x) = x^3 + x^2 - 12x?

Nov 2, 2016

$f \left(x\right)$ has zeros $x = 0$, $x = - 4$ and $x = 3$

#### Explanation:

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

Note that $4 \cdot 3 = 12$ and $4 - 3 = 1$

Hence we find:

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

So $f \left(x\right)$ has zeros $x = 0$, $x = - 4$ and $x = 3$