Question

How do you factor `5x^4+15x^3-30x^2-60x`?

Answer 1

`5x^4+15x^3-30x^2-60x = 5x(x^3+3x^2-6x-12)`

Explanation:

Given:

`5x^4+15x^3-30x^2-60x`

First note that all of the terms are divisible by `5x`, so we can separate that out as a factor first to find:

`5x^4+15x^3-30x^2-60x = 5x(x^3+3x^2-6x-12)`

By the rational roots theorem, any rational zeros of the remaining cubic 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 `1` of the leading term.

That means that the only possible rational zeros are:

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

We find:

`(color(blue)(2))^3+3(color(blue)(2))^2-6(color(blue)(2))-12 = 8+12-12-12 = -4 < 0`

`(color(blue)(3))^3+3(color(blue)(3))^2-6(color(blue)(3))-12 = 27+27-18-12 = 24 > 0`

So there is an irrational zero in `(2, 3)`

In fact all of the zeros of this cubic are real but irrational.

It is possible to find them in terms of irreducible cube roots of complex numbers or in terms of trigonometric functions and thus get a factorisation of the form `(x-alpha)(x-beta)(x-gamma)`, but it probably beyond the scope of your course.

So let's stick with the rational factorisation:

`5x^4+15x^3-30x^2-60x = 5x(x^3+3x^2-6x-12)`