Factoring out $4x^3+2x^2y-2xy^2-y^3$

Question

This can be factored as follows:

$$4x^3+2x^2y-2xy^2-y^3 = (2x^2-y^2)(2x+y)$$

What is a systematic way for finding this factorization?

Answer 1

\begin{gather*} 4x^{3} +2x^{2} y-2xy^{2} -y^{3}\\ =\left( 4x^{3} -2xy^{2}\right) +\left( 2x^{2} y-y^{3}\right)\\ =2x\left( 2x^{2} -y^{2}\right) +y\left( 2x^{2} -y^{2}\right)\\ =( 2x+y)\left( 2x^{2} -y^{2}\right) \end{gather*} Does this help?

Answer 2

Hint

$$4x^3-2xy^2=2x(2x^2-y^2)$$

$$2x^2y-y^3=y(2x^2-y^2)$$

Can you take it from here?

Answer 3

The systematic way here is whenever you have a sum of 4 terms, you can check to see if factor by grouping works. In a sum of the form $a+b+c+d$, factor by grouping works if and only if $ad=bc$. If it does, then you pull out the greatest common factor of $a$ and $b$. When you do so, you will be GUARANTEED to be able to pull a factor out of $c+d$ to match the binomial you got when factoring out $a+b$, as in lab's answer above.