Question

How do you factor the trinomial `15x^3 - 87x^2 - 264x`?

Answer 1

`15x^3-87x^2-264x = 3x(5x+11)(x-8)`

Explanation:

All of the terms are divisible by `3x`, so separate that out as a factor first:

`15x^3-87x^2-264x = 3x(5x^2-29x-88)`

Next use an AC method to factor `5x^2-29x-88`:

Look for a pair of factors of `AC=5*88 = 440` which differ by `B=29`.

The pair `40, 11` works.

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

`5x^2-29x-88`

`=5x^2-40x+11x-88`

`=(5x^2-40x)+(11x-88)`

`=5x(x-8)+11(x-8)`

`=(5x+11)(x-8)`

Putting it all together:

`15x^3-87x^2-264x = 3x(5x+11)(x-8)`

Answer 2

3x(5x + 11)(x - 8)

Explanation:

`f(x) = 3xy = 3x(5x^2 - 29x - 88)`.
Factor the trinomial y in parentheses.
`y = 5x^2 - 29x - 88 =` 5(x + p)(x + q).
Converted trinomial: `y' = x^2 - 29x - 440 =` (x + p')(x + q')
p' and q' have opposite signs because ac < 0.
Factor pairs of (ac = - 440) --> ... (10, -44) (11, -40). This last sum is (-29 = b). Then, p' = 11 and q' = -40.
Back to original trinomial y, `p = (p')/a = 11/5` and `q = (q')/a = -40/5 = -8`.
Factored form --> `y = 5(x + 11/5)(x - 8) = (5x + 11)(x - 8)`
Factored form --> `f(x) = 3xy = 3x(5x + 11)(x - 8)`