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

Aug 7, 2017

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

#### Explanation:

Factor:

$\left(4 {x}^{3} - {x}^{2}\right) \left(16 x - 4\right)$

Factor out the greatest common factor ${x}^{2}$ from the first binomial.

${x}^{2} \left(4 x - 1\right) \left(16 x - 4\right)$

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

$4 {x}^{2} \left(4 x - 1\right) \left(4 x - 1\right)$

Simplify.

$4 {x}^{2} {\left(4 - 1\right)}^{2}$

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

Look at $\boldsymbol{4 {x}^{3} - {x}^{2}}$. There is an ${x}^{2}$ term we can factor out:

$\left(4 {x}^{3} - {x}^{2}\right) = {x}^{2} \left(4 x - 1\right)$

And now let's look at $\boldsymbol{16 x - 4}$. We can factor out a 4 from each term:

$\left(16 x - 4\right) = 4 \left(4 x - 1\right)$

This now gives us:

$\left(4 {x}^{3} - {x}^{2}\right) \left(16 x - 4\right) = {x}^{2} \left(4 x - 1\right) 4 \left(4 x - 1\right)$

which we can rewrite as:

$4 {x}^{2} {\left(4 x - 1\right)}^{2}$