# How do you factor completely 5x^3 - 45x?

Apr 26, 2016

$5 {x}^{3} - 45 x = 5 x \left(x - 3\right) \left(x + 3\right)$

#### Explanation:

Notice that both of the terms are divisible by $5 x$, so we can separate that out as a factor first, then use the difference of squares identity:

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

with $a = x$ and $b = 3$ as follows:

$5 {x}^{3} - 45 x = 5 x \left({x}^{2} - 9\right) = 5 x \left({x}^{2} - {3}^{2}\right) = 5 x \left(x - 3\right) \left(x + 3\right)$