# How do you factor F(x)= -x^4 + 2x^3 + 4x^2 - 8x?

Aug 3, 2015

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

#### Explanation:

Notice that you can group the first two terms and the second two terms together and factor them to get

$F \left(x\right) = \left(- {x}^{4} + 2 {x}^{3}\right) + \left(4 {x}^{2} - 8 x\right)$

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

You can factor these two terms by $\left(x - 2\right)$ to get

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

Notice that you can factor the second term of the expression by $x$

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

Finally, you can factor $\left(4 - {x}^{2}\right)$ as the difference of two squares

$\textcolor{b l u e}{{a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)}$

$4 - {x}^{2} = {2}^{2} - {x}^{2} = \left(2 - x\right) \left(2 + x\right)$
$F \left(x\right) = \left(x - 2\right) \cdot x \cdot \left(2 - x\right) \cdot \left(2 + x\right)$
$F \left(x\right) = - x \left(x - 2\right) \cdot \left(x - 2\right) \cdot \left(2 + x\right) = \textcolor{g r e e n}{- x {\left(x - 2\right)}^{2} \left(x + 2\right)}$