# How do you factor by grouping 12x^2-7x-5?

Jun 6, 2015

The trick is to find how to split the middle $- 7 x$ term.

The split comes from a pair of factors of $A C = 12 \cdot 5 = 60$ whose difference is $B = 7$. The pair $12 , 5$ works.

$12 {x}^{2} - 7 x - 5$

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

$= \left(12 {x}^{2} - 12 x\right) + \left(5 x - 5\right)$

$= 12 x \left(x - 1\right) + 5 \left(x - 1\right)$

$= \left(12 x + 5\right) \left(x - 1\right)$

Jun 6, 2015

When a + b + c = 0, you are advised to use the Shortcut.
One factor is (x - 1) and the other is (-c/a = 5/12).

y = (x - 1)( 12x + 5)