Question

How do you factor `15x^3+7x^2-40x+12`?

Answer 1

`15x^3+7x^2-40x+12 = (3x-1)(5x-6)(x+2)`

Explanation:

`f(x) = 15x^3+7x^2-40x+12`

By the rational root theorem, any rational zeros of `f(x)` are expressible in the form `p/q` for integers `p`, `q` with `p` a divisor of the constant term `12` and `q` a divisor of the coefficient `15` of the leading term.

That means that the only possible rational zeros are:

`+-1/15`, `+-2/15`, `+-1/5`, `+-4/15`, `+-1/3`, `+-2/5`, `+-2/3`, `+-3/5`, `+-4/5`, `+-1`, `+-6/5`, `+-4/3`, `+-2`, `+-12/5`, `+-3`, `+-4`, `+-6`, `+-12`

Trying each in turn, the first one that works is `x=1/3`:

`f(1/3) = 15/27+7/9-40/3+12 = (5+7-120+108)/9 = 0`

So `(3x-1)` is a factor:

`15x^3+7x^2-40x+12 = (3x-1)(5x^2+4x-12)`

To factor the remaining quadratic, use an AC method.

Find a pair of factors of `AC = 5*12 = 60` which differ by `B=4`

The pair `10, 6` works.

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

`5x^2+4x-12`

`=5x^2+10x-6x-12`

`=(5x^2+10x)-(6x+12)`

`=5x(x+2)-6(x+2)`

`=(5x-6)(x+2)`

Putting it all together:

`15x^3+7x^2-40x+12 = (3x-1)(5x-6)(x+2)`