Factor completely #(4x^3-x^2)(16x-4)#?

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Answer 1 · 2017-08-07T01:03:19

#(4x^3-x^2)(16x-4)=color(blue)(4x^2(4-1)^2#

Factor:

#(4x^3-x^2)(16x-4)#

Factor out the greatest common factor #x^2# from the first binomial.

#x^2(4x-1)(16x-4)#

Factor out the greatest common factor #4# from the second binomial.

#4x^2(4x-1)(4x-1)#

Simplify.

#4x^2(4-1)^2#

Answer 2 · 2017-08-07T01:09:51

This is already in a factored form, but there are certainly more factors that can be taken.

Look at #bb(4x^3-x^2)#. There is an #x^2# term we can factor out:

#(4x^3-x^2)=x^2(4x-1)#

And now let's look at #bb(16x-4)#. We can factor out a 4 from each term:

#(16x-4)=4(4x-1)#

This now gives us:

#(4x^3-x^2)(16x-4)=x^2(4x-1)4(4x-1)#

which we can rewrite as:

#4x^2(4x-1)^2#