# How do you factor completely: 10x^2 + 2x − 8?

Jul 12, 2015

Factor: $y = 10 {x}^{2} + 2 x - 8$

#### Explanation:

In this case, use the shortcut, since (a - b + c) = 0. One factor is (x + 1) and the other is (x + c/a = - 8/10).
y = (x + 1)(10x - 8) = 2(x + 1)(5x - 4)

You may also use the new AC Method.
Converted $y ' = {x}^{2} + 2 x - 80 = \left(x - p '\right) \left(x - q '\right) .$
Factor pairs of (-80) -> ...(-4, 20)(-8, 10). This sum is 2 = b. Then p' = -8 and q' = 10.
Then, $p = \frac{p '}{a} = - - \frac{8}{10}$ and $q = \frac{q '}{a} = \frac{10}{10} = 1.$

y = 10(x - 8/10)(x + 1) = (10x - 8)(x + 1)= 2(5x - 4)(x + 1)