Question

How do you factor `x^3-x^2-14x+24` ?

Answer 1

`x^3-x^2-14x+24 = (x-2)(x+4)(x-3)`

Explanation:

Given:

`f(x) = x^3-x^2-14x+24`

By the rational roots theorem, any rational zeros of this polynomial are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `24` and `q` a divisor of the coefficient `1` of the leading term.

So the only possible rational zeros are:

`+-1, +-2, +-3, +-4, +-6, +-8, +-12, +-24`

We find:

`f(2) = color(blue)(2)^3-color(blue)(2)^2-14(color(blue)(2))+24 = 8-4-28+24 = 0`

So `x=2` is a zero and `(x-2)` a factor:

`x^3-x^2-14x+24 = (x-2)(x^2+x-12)`

To factor the remaining quadratic, find a pair of factors of `12` which differ by `1`. The pair `4, 3` works.

Hence:

`x^2+x-12 = (x+4)(x-3)`

So:

`x^3-x^2-14x+24 = (x-2)(x+4)(x-3)`