Question

How do you factor `y= x^3-5x^2-2x+24` ?

Answer 1

Use the rational root theorem to get started, then factor the remaining quadratic to find:

`x^3-5x^2-2x+24 = (x+2)(x-4)(x-3)`

Explanation:

Let `f(x) = x^3-5x^2-2x+24`

By the rational root theorem, any rational zeros of `f(x)` must be expressible in the for `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.

That means that the only possible rational zeros are the factors of `24`, namely:

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

Try each in turn:

`f(1) = 1-5-2+24 = 18`

`f(-1) = -1-5+2+24 = 20`

`f(2) = 8-20-4+24 = 8`

`f(-2) = -8-20+4+24 = 0`

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

`x^3-5x^2-2x+24 = (x+2)(x^2-7x+12)`

We can factor `x^2-7x+12` by noting that `4xx3 = 12` and `4+3=7`, so:

`x^2-7x+12 = (x-4)(x-3)`

Putting it all together:

`x^3-5x^2-2x+24 = (x+2)(x-4)(x-3)`