Question

How do you factor the trinomial `16a^2-22ab-3b^2`?

Answer 1

Use an AC Method: Look for a pair of factors of `16xx3 = 48` with difference `22`, hence find:

`16a^2-22ab-3b^2 = (8a+b)(2a-3b)`

Explanation:

Multiply the first coefficient `A=16` by the third coefficient (ignoring sign) `C=3` to get `48`.

Then since the sign of the third coefficient is negative, look for a pair of factors of `AC=48` whose difference is `B=22`.

The pair `(B1, B2) = (24, 2)` works, so use that to split the middle term:

`16a^2-24ab+2ab-3b^2`

We can then factor by grouping to find:

`16a^2-24ab+2ab-3b^2`

`=(16a^2-24ab)+(2ab-3b^2)`

`=8a(2a-3b)+b(2a-3b)`

`=(8a+b)(2a-3b)`

Alternatively, just write down the pairs `(A, B1)` and `(A, B2)` with appropriate signs and divide by highest common factors to find pairs of coefficients as follows:

`(A, B1) -> (16, -24) -> (2, -3) -> (2a-3b)` (dividing through by `8`)

`(A, B2) -> (16, 2) -> (8, 1) -> (8a+b)` (dividing through by `2`)

The slight complication is the choosing of signs for the second term so that the resulting `+-B1 +- B2 = -22`