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

${x}^{2} \left(6 x - 5\right) \left(3 x + 4\right)$

#### Explanation:

$18 {x}^{4} + 9 {x}^{3} - 20 {x}^{2}$

We can first factor out ${x}^{2}$ to get:

${x}^{2} \left(18 {x}^{2} + 9 x - 20\right)$

From here, we need to find factors for the terms within the brackets in the form of $\left(a x + b\right) \left(c x + d\right)$ such that:

$a c = 18$
$b d = - 20$
$a d + b c = 9$

and there's some trial and error to this.

Let's play with $a c = 18$. 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 $b d = - 20$ 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:

${x}^{2} \left(6 x - 5\right) \left(3 x + 4\right)$