How do you factor $3 a^{2} + 4 a - 4$ $3a^2 + 4a - 4$ ?

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How do you factor $3 a^{2} + 4 a - 4$ $3a^2 + 4a - 4$ ?

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Answer 1 · 2016-04-16T07:02:36

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

Explanation:

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

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

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

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

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

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

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

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

$= ((3 a + 2) - 4) ((3 a + 2) + 4)$ $=((3a+2)-4)((3a+2)+4)$

$= (3 a - 2) (3 a + 6)$ $=(3a-2)(3a+6)$

$= 3 (3 a - 2) (a + 2)$ $=3(3a-2)(a+2)$

So dividing by $3$ $3$ we have:

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

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