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

1 Answer
MathFact-orials.blogspot.com
Sep 1, 2016

#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)#

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