How do you factor $3 x^{2} + 5 x - 12$ $3x^2 + 5x - 12$ ?

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Answer 1 · 2015-05-30T19:00:24

Use a version of the AC Method.

$A = 3$ $A=3$ , $B = 5$ $B=5$ , $C = 12$ $C=12$

Since the sign on the constant term is $-$ $-$ , look for a pair of factors of $A C = 36$ $AC=36$ whose difference is $B = 5$ $B=5$

The pair $4$ $4$ , $9$ $9$ works.

Now use that pair to split the middle term, then factor by grouping:

$3 x^{2} + 5 x - 12$ $3x^2+5x-12$

$= 3 x^{2} - 4 x + 9 x - 12$ $=3x^2-4x+9x-12$

$= (3 x^{2} - 4 x) + (9 x - 12)$ $=(3x^2-4x)+(9x-12)$

$= x (3 x - 4) + 3 (3 x - 4)$ $=x(3x-4)+3(3x-4)$

$= (x + 3) (3 x - 4)$ $=(x+3)(3x-4)$

How do you factor 3x^2 + 5x - 123x2+5x−123x^2 + 5x - 12?