How do you factor 16x^2-x^2y^4?

May 12, 2016

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

Explanation:

First, notice that both terms have a common factor of ${x}^{2}$, so we can factor it out:

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

Focusing on just the $\left(16 - {y}^{4}\right)$, notice that both of these are squared terms:

• $16 = {4}^{2}$
• ${y}^{4} = {\left({y}^{2}\right)}^{2}$

This will be useful since $\left(16 - {y}^{4}\right)$ is a difference of squares, which can be factored as: ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$. Thus, we see that:

${x}^{2} \left(16 - {y}^{4}\right) = {x}^{2} \left({4}^{2} - {\left({y}^{2}\right)}^{2}\right) = {x}^{2} \left(4 + {y}^{2}\right) \left(4 - {y}^{2}\right)$

Note that $\left(4 - {y}^{2}\right)$ is also a difference of squares, since ${y}^{2}$ is obviously squared and $4 = {2}^{2}$. We see that:

${x}^{2} \left(4 + {y}^{2}\right) \left(4 - {y}^{2}\right) = {x}^{2} \left(4 + {y}^{2}\right) \left({2}^{2} - {y}^{2}\right) = {x}^{2} \left(4 + {y}^{2}\right) \left(2 + y\right) \left(2 - y\right)$