# How do you factor 3a^2 + 4a - 4?

Apr 16, 2016

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

#### Explanation:

We can factor this by completing the square, then using the difference of squares identity:

${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$

with $A = \left(3 a + 2\right)$ and $B = 4$, as follows:

To make the arithmetic a little easier, first multiply by $3$ (making the leading term into a square), remembering to divide by $3$ at the end:

$3 \left(3 {a}^{2} + 4 a - 4\right)$

$= 9 {a}^{2} + 12 a - 12$

$= {\left(3 a + 2\right)}^{2} - 4 - 12$

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

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

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

$= 3 \left(3 a - 2\right) \left(a + 2\right)$

So dividing by $3$ we have:

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