How do you find all the zeros of f(x)=x^3 +6x^2 - 6x-36?

Mar 5, 2016

Factor $f \left(x\right)$ to see that the zeroes occur at $- 6$, $\sqrt{6}$, and $- \sqrt{6}$

Explanation:

Using the technique of factoring by grouping as well as the difference of squares formula , we can factor $f \left(x\right)$ as

${x}^{3} + 6 {x}^{2} - 6 x - 36 = {x}^{2} \left(x + 6\right) - 6 \left(x + 6\right)$

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

$= \left(x + \sqrt{6}\right) \left(x - \sqrt{6}\right) \left(x + 6\right)$

In its factored form, we can see that the zeros occur when any of the factors become $0$, that is, when $x = - 6$ or $x = \pm \sqrt{6}$.