# How do you factor 12a^2 + 7ab - 10b^2?

May 21, 2015

Since this quadratic is homogeneous (that is all of the terms have the same order, viz 2), this problem is equivalent to the problem of factoring $12 {x}^{2} + 7 x - 10$.

$12 {x}^{2} + 7 x - 10$ is of the form $a {x}^{2} + b x + c$, with $a = 12$, $b = 7$ and $c = - 10$ (not to be confused with the $a$ and $b$ in your original problem).

This has discriminant given by the formula:

$\Delta = {b}^{2} - 4 a c = {7}^{2} - \left(4 \times 12 \times - 10\right) = 49 + 480$

$= 529 = {23}^{2}$

Since this is a perfect square, the polynomial equation $12 {x}^{2} + 7 x - 10 = 0$ has rational roots, given by:

$x = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{- 7 \pm 23}{24}$

that is $x = - \frac{30}{24} = - \frac{5}{4}$ or $x = \frac{16}{24} = \frac{2}{3}$.

This allows us to deduce that:

$12 {x}^{2} + 7 x - 10 = \left(4 x + 5\right) \left(3 x - 2\right)$

and hence that:

$12 {a}^{2} + 7 a b - 10 {b}^{2} = \left(4 a + 5 b\right) \left(3 a - 2 b\right)$