Question

How do you factor `28x^2+5xy-12y^2`?

Answer 1

Use the quadratic formula to find factors:

`28x^2+5xy-12y^2 = (4x+3y)(7x-4y)`

Explanation:

If you divide this polynomial by `y^2` then you get a quadratic in `x/y`:

`(28x^2+5xy-12y^2) / y^2 = 28(x/y)^2 + 5(x/y) - 12`

So if we let `t = x/y` we get a quadratic in `t`

`28t^2+5t-12`

which is of the form `at^2+bt+c`, with `a = 28`, `b=5` and `c=-12`.

This has zeros for values of `t` given by the quadratic formula:

`t = (-b+-sqrt(b^2-4ac))/(2a) = (-5+-sqrt(5^2-(4xx28xx-12)))/(2*28)`

`=(-5+-sqrt(25+1344))/56 = (-5+-sqrt(1369))/56 = (-5+-37)/56`

That is `t = -42/56 = -3/4` and `t = 32/56 = 4/7`

Hence:

`28t^2+5t-12 = (4t+3)(7t-4)`

So

`28(x/y)^2+5(x/y)-12 = (4(x/y)+3)(7(x/y)-4)`

Then multiply through by `y^2` to find:

`28x^2+5xy-12y^2 = (4x+3y)(7x-4y)`