Question

How do you factor `7n^2-13n-2`?

Answer 1

`7(n - 2)(n + 1/7)`

Explanation:

`7n^2 - 13n - 2 = 0 \Rightarrow n = (13 ± sqrt{225})/14`

`n_1 = (13 + 15)/14 = 2`

`n_2 = (13 - 15)/14 = -1/7`

`ax^2 + bx + c = a(x - x_1)(x - x_2)`

Answer 2

Alternatively, use a variant of the AC Method to find:

`7n^2-13n-2 = (7n+1)(n-2)`

Explanation:

Given `7n^2-13n-2`, let `A=7`, `B=13` and `C=2`.

Notice that the sign of the constant term is negative, so look for a pair of factors of `AC = 7*2 = 14`, whose difference is `B=13`.

It should not take long to spot that the pair `B1=1, B2=14` works.

Next, for each of the pairs `(A, B1)` and `(A, B2)` divide by the HCF (highest common factor) to find a pair of coefficients of a factor, choosing signs as appropriate:

`(7, 1) -> (7, 1) -> (7n+1)`
`(7, 14) -> (1, 2) -> (n-2)`

We choose `+` for the pair `(A, B1)` and `-` for the pair `(A, B2)` to get `(+B1)+(-B2) = -13` - the coefficient of the middle term.