# How do you factor completely 8w^3 - 4w^2 - 60w?

Jul 31, 2016

$4 w \left(2 w + 5\right) \left(w - 3\right)$

#### Explanation:

Always look for a common factor first. In this case it is $4 w$

$8 {w}^{3} - 4 {w}^{2} - 60 w = 4 w \left(2 {w}^{2} - w - 15\right)$

Now, find factors of 2 and 15 which subtract to give 1.
There must be more negatives.

find the cross products and subtract.
$\text{ 2 5 } \Rightarrow 1 \times 5 = 5$
$\text{ 1 3 "rArr2 xx3 = 6" 6-5 = 1}$

But we need $- 1 , \text{ 6 must be negative and 5 must be positive}$

=$4 w \left(2 w + 5\right) \left(w - 3\right)$