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

Jun 4, 2016

Factor by grouping to find zeros:

$x = 2$, $x = - 2$ and $x = 1$

#### Explanation:

Notice that the ratio of the first and second terms is the same as that between the third and fourth terms. So this cubic can be factored by grouping:

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

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

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

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

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

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

Hence zeros:

$x = 2$, $x = - 2$ and $x = 1$