# How do you find the zeros of x^3+x^2-4x-4?

Aug 19, 2016

This cubic has zeros: $2 , - 2 , - 1$

#### Explanation:

This cubic factors by grouping then using the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = x$ and $b = 2$ as follows:

${x}^{3} + {x}^{2} - 4 x - 4$

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

$= {x}^{2} \left(x + 1\right) - 4 \left(x + 1\right)$

$= \left({x}^{2} - 4\right) \left(x + 1\right)$

$= \left({x}^{2} - {2}^{2}\right) \left(x + 1\right)$

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

Hence zeros:

$2 , - 2 , - 1$