Question

How do you factor `12c^2+23c-9`?

Answer 1

`12c^2+23c-9 = (3c-1)(4c+9)`

Explanation:

`12c^2+23c-9`

Use an AC method:

Find a pair of factors of `AC = 12*9 = 108 = 2^2*3^3` which differ by `23`

Since `23` is not divisible by `3` or `2`, one of the factors is divisible by `3^3 = 27` and one factor (possibly the same one) is divisible by `2^2 = 4`. Note that `27-4 = 23`, so we have found the required pair in `27, 4`

Use this pair to split the middle term and factor by grouping:

`12c^2+23c-9 = (12c^2+27c)-(4c+9)`

`color(white)(12c^2+23c-9) = 3c(4c+9)-1(4c+9)`

`color(white)(12c^2+23c-9) = (3c-1)(4c+9)`