Question

How do you factor `3x^3 - 9x^2 - 54x + 120`?

Answer 1

`3x^3-9x^2-54x+120 = 3(x+2)(x-5)(x+4)`

Explanation:

`3x^3-9x^2-54x+120 = 3(x^3-3x^2-18x+40)`

By the rational roots theorem, any rational zeros of `x^3-3x^2-18x+40` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `40` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1, +-2, +-4, +-5, +-8, +-10, +-20, +-40`

Trying each in turn, with `x=2` we get:

`x^3-3x^2-18x+40=8-3(4)-18(2)+40 = 8-12-36+40 = 0`

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

`x^3-3x^2-18x+40 = (x-2)(x^2-x-20)`

To factor the remaining quadratic, note that `5-4=1` and `5*4=20`.

Hence:

`x^2-x-20 = (x-5)(x+4)`

Putting it all together:

`3x^3-9x^2-54x+120 = 3(x+2)(x-5)(x+4)`