# How do you factor completely 12x^3 - 3xy^2?

Feb 19, 2017

$3 x \left(2 x + y\right) \left(2 x - y\right)$

#### Explanation:

$12 {x}^{3} - 3 x {y}^{2}$

take out all common factors

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

use difference of squares

$3 x \left(2 x + y\right) \left(2 x - y\right)$

Feb 19, 2017

To factor, we have to divide out common factors in each term. Let's first write out the factors:

$12 {x}^{3}$ = $4 \cdot 3 \cdot x \cdot x \cdot x$

$- 3 x {y}^{2}$ = $3 \cdot - 1 \cdot x \cdot y \cdot y$

Now look for the factors that each term share. We see that there is a $3$ and an $x$ that is found in each list. Now we have to take this out of each list.

$\left(3 \cdot x\right) \left(4 \cdot x \cdot x\right)$

$\left(3 \cdot x\right) \left(- 1 \cdot y \cdot y\right)$

Remember that these were being subtracted, so rewrite:

$\left(3 \cdot x\right) \left[\left(4 \cdot x \cdot x\right) - \left(1 \cdot y \cdot y\right)\right]$

Simplify.

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

Use difference of squares to simplify further.

$\left(3 x\right) \left(2 x - y\right) \left(2 x + y\right)$