# How do you find all the zeros of x^3-12x^2+12x+80 with -2 as a zero?

Jul 14, 2016

Zeros: -2, 10, 4.

#### Explanation:

Since we are told that $x = - 2$ is a zero, this cubic has a factor $\left(x + 2\right)$:

${x}^{3} - 12 {x}^{2} + 12 x + 80 = \left(x + 2\right) \left({x}^{2} - 14 x + 40\right)$

To factor the remaining quadratic, we can find a pair of factors of $40$ with sum $14$. The pair $10 , 4$ works, hence:

${x}^{2} - 14 x + 40 = \left(x - 10\right) \left(x - 4\right)$

So the other two zeros are $x = 10$ and $x = 4$