How do you factor  25x^4 + 16x^2y^2 + 4y^4?

May 10, 2016

$25 {x}^{4} + 16 {x}^{2} {y}^{2} + 4 {y}^{4} = \left(5 {x}^{2} - 2 x y + 2 {y}^{2}\right) \left(5 {x}^{2} + 2 x y + 2 {y}^{2}\right)$

Explanation:

Note that:

$\left({a}^{2} - k a b + {b}^{2}\right) \left({a}^{2} + k a b + {b}^{2}\right) = {a}^{4} + \left(2 - {k}^{2}\right) {a}^{2} {b}^{2} + {b}^{4}$

Putting $a = \sqrt{5} x$ and $b = \sqrt{2} y$ we get:

$\left(5 {x}^{2} - k \sqrt{10} x y + 2 {y}^{2}\right) \left(5 {x}^{2} + k \sqrt{10} x y + 2 {y}^{2}\right)$

$= 25 {x}^{4} + 10 \left(2 - {k}^{2}\right) {x}^{2} {y}^{2} + 4 {y}^{4}$

Solve:

$16 = 10 \left(2 - {k}^{2}\right) = 20 - 10 {k}^{2}$

So:

$10 {k}^{2} = 20 - 16 = 4$

${k}^{2} = \frac{2}{5} = \frac{10}{25}$

$k = \pm \frac{\sqrt{10}}{5}$

So:

$k \sqrt{10} = \pm \frac{\sqrt{10}}{5} \cdot \sqrt{10} = \pm 2$

Hence:

$25 {x}^{4} + 16 {x}^{2} {y}^{2} + 4 {y}^{4} = \left(5 {x}^{2} - 2 x y + 2 {y}^{2}\right) \left(5 {x}^{2} + 2 x y + 2 {y}^{2}\right)$

May 10, 2016

$\left(5 {x}^{2} + 2 x y + 2 {y}^{2}\right) \left(5 {x}^{2} - 2 x y + 2 {y}^{2}\right)$

Explanation:

This is kind of tricky and involves some intuition. First, notice that we have a lot of square terms, e.g. $25 {x}^{4} = {\left(5 {x}^{2}\right)}^{2} , 16 {x}^{2} {y}^{2} = {\left(4 x y\right)}^{2} ,$ and $4 {y}^{4} = {\left(2 {y}^{2}\right)}^{2}$.

A good trick for these is to try to create squared trinomials, which come in the form ${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$. If we let ${a}^{2} = 25 {x}^{4}$, then $a = 5 {x}^{2}$, and if ${b}^{2} = 4 {y}^{4}$, then $b = 2 {y}^{2}$. Thus, $2 a b = 20 {x}^{2} {y}^{2}$.

From this, we can see that we almost have ${\left(5 {x}^{2} + 2 {y}^{2}\right)}^{2} = 25 {x}^{4} + 20 {x}^{2} {y}^{2} + 4 {y}^{4}$, but there is a discrepancy between the $20 {x}^{2} {y}^{2}$ and $16 {x}^{2} {y}^{2}$ terms.

We can write the given expression as:

$25 {x}^{4} + 16 {x}^{2} {y}^{2} + 4 {y}^{4} = 25 {x}^{4} + \left(20 {x}^{2} {y}^{2} - 4 {x}^{2} {y}^{2}\right) + 4 {y}^{4}$

We can then reorder this so that $25 {x}^{4} + 20 {x}^{2} {y}^{2} + 4 {y}^{4}$ is present:

$25 {x}^{4} + \left(20 {x}^{2} {y}^{2} - 4 {x}^{2} {y}^{2}\right) + 4 {y}^{4} = \left(25 {x}^{4} + 20 {x}^{2} {y}^{2} + 4 {y}^{4}\right) - 4 {x}^{2} {y}^{2}$

Recall that $25 {x}^{4} + 20 {x}^{2} {y}^{2} + 4 {y}^{4} = {\left(5 {x}^{2} + 2 {y}^{2}\right)}^{2}$. Also, it will be important that $4 {x}^{2} {y}^{2} = {\left(2 x y\right)}^{2}$.

$\left(25 {x}^{4} + 20 {x}^{2} {y}^{2} + 4 {y}^{4}\right) - 4 {x}^{2} {y}^{2} = {\left(5 {x}^{2} + 2 {y}^{2}\right)}^{2} - {\left(2 x y\right)}^{2}$

What we now have is a difference of squares, which can be factored as ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$. Here, $a = 5 {x}^{2} + 2 {y}^{2}$ and $b = 2 x y$. Applying this, we obtain:

${\left(5 {x}^{2} + 2 {y}^{2}\right)}^{2} - {\left(2 x y\right)}^{2} = \left(5 {x}^{2} + 2 x y + 2 {y}^{2}\right) \left(5 {x}^{2} - 2 x y + 2 {y}^{2}\right)$