Question

How do you factor completely `18x^4+9x^3-20x^2`?

Answer 1

`x^2(6x-5)(3x+4)`

Explanation:

Let's start with the original:

`18x^4+9x^3-20x^2`

We can first factor out `x^2` to get:

`x^2(18x^2+9x-20)`

From here, we need to find factors for the terms within the brackets in the form of `(ax+b)(cx+d)` such that:

`ac=18`
`bd=-20`
`ad + bc = 9`

and there's some trial and error to this.

Let's play with `ac=18`. We can have as factors (18,1), (9,2), (6,3), and then those sets with all the numbers negative (for a total of 6 possibilities).

We have a different story with `bd=-20` in terms of signage - one term must be negative. So we can have (20,-1), (10, -2), (5, -4) and then those sets with the signs reversed.

So let's see if we can find a combo that works:

For a=6, b=-4, c=3, d=5; ac = 18, bd = -20, ad+bc = 30-12=18
For a=6, b=-5, c=3, d=4; ac = 18, bd=-20, ad+bc = 24-15=9

Ok - found a combo that works. Let's factor:

`x^2(6x-5)(3x+4)`