# How do you factor 4(x+3)^2-9(x-1)^2?

Feb 14, 2016

This can be factored as a difference of squares.

#### Explanation:

Differences of squares are of the form ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$. We can find the factors by finding the square roots of each part of the expression:

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

$\sqrt{9 {\left(x - 1\right)}^{2}} = 3 \left(x - 1\right)$

Thus, the expression can be factored into:

$= \left(2 \left(x + 3\right) - 3 \left(x - 1\right)\right) \left(2 \left(x + 3\right) + 3 \left(x - 1\right)\right)$

$= \left(2 x + 6 - 3 x + 3\right) \left(2 x + 6 + 3 x - 3\right)$

$= \left(9 - x\right) \left(3 + 5 x\right)$

Hopefully this helps!