Question

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

Answer 1

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 `12x^2+7x-10`.

`12x^2+7x-10` is of the form `ax^2+bx+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^2-4ac = 7^2-(4xx12xx-10) = 49+480`

`= 529 = 23^2`

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

`x = (-b+-sqrt(Delta))/(2a) = (-7+-23)/24`

that is `x=-30/24=-5/4` or `x = 16/24 = 2/3`.

This allows us to deduce that:

`12x^2+7x-10 = (4x+5)(3x-2)`

and hence that:

`12a^2+7ab-10b^2 = (4a+5b)(3a-2b)`