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:

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

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