Question

How do you factor completely: `18x^2 − 21x − 15`?

Answer 1

`18x^2-21x-15 = 3(6x^2-7x-5) = 3(2x+1)(3x-5)`

Explanation:

First notice the common scalar factor `3` and separate that out:

`18x^2-21x-15 = 3(6x^2-7x-5)`

Then factor `6x^2-7x-5` using a variant of the AC Method.

Let `A=6`, `B=7`, `C=5`.

Look for a pair of factors of `AC=30` whose difference is `B=7`.

The pair `B1=10`, `B2=3` works.

Next for each of the pairs `(A, B1)`, `(A, B2)` divide by the HCF (highest common factor) to give a pair of coefficients of a linear factor, choosing suitable signs:

`(A, B1) = (6, 10) -> (3, 5) -> (3x-5)`
`(A, B2) = (6, 3) -> (2, 1) -> (2x+1)`

Hence:

`18x^2-21x-15 = 3(6x^2-7x-5) = 3(2x+1)(3x-5)`