How do you factor completely #12x^3-18x^2-8x+12#?

1 Answer
MathFact-orials.blogspot.com
May 17, 2016

In problems like this, look for the first 2 terms and the second 2 terms to have a common factor - this one factored completely results in
#2(2x-3)(3x^2-2)#

Explanation:

In a problem like this one, we look for the (generally first two) terms and the second two terms and find ways to factor each of them to create a common factor to them both.

In this problem, we can see that the first two terms have a similar pattern to them as do the second two terms. And in fact, once we start factoring, there will be a common factor to them both.

Let's take the first 2 terms: #12x^3 - 18x^2#. We can factor out an #x^2# and also a 6, which leaves us with

#(6x^2)(2x-3)#

Can we factor something out of the second 2 terms, #-8x+12# to get us to #2x-3#? Yes we can - factor out a #-4# from each term. Which will give us

#(-4)(2x-3)#

So right now we have

#(6x^2)(2x-3)+(-4)(2x-3)#

We can take the common factor #2x-3# and factor that out and end up with

#(2x-3)(6x^2-4)#

Let's factor out a 2 from the #6x^2-4# term, which gives us the final answer of

#2(2x-3)(3x^2-2)#

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