# What are all the zeroes of f(x) = 2x^3 - 2x^2 - 8x + 8?

Dec 13, 2015

$x = - 2 , 1 , 2$

#### Explanation:

Factor the polynomial.

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

Factor by grouping.

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

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

Recognize that ${x}^{2} - 4$ is a difference of squares.

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

Now, to find the zeros of a function, find the times when $f \left(x\right) = 0$.

$0 = 2 \left(x + 2\right) \left(x - 2\right) \left(x - 1\right)$

When there is a product of things that equal $0$, at least one of the terms MUST equal $0$. To find when this happens, set each non-constant term equal to $0$.

$x + 2 = 0$
$\textcolor{b l u e}{x = - 2}$

$x - 2 = 0$
$\textcolor{b l u e}{x = 2}$

$x - 1 = 0$
$\textcolor{b l u e}{x = 1}$

A good way to double check is by checking a graph.
graph{2x^3-2x^2-8x+8 [-7.96, 12.04, -3.56, 6.44]}

The zeros, when the graph crosses the $x$-axis, occur at $x = - 2 , 1 , 2$.