Question

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

Answer 1

`25x^4+16x^2y^2+4y^4 = (5x^2-2xy+2y^2)(5x^2+2xy+2y^2)`

Explanation:

Note that:

`(a^2-kab+b^2)(a^2+kab+b^2) = a^4+(2-k^2)a^2b^2+b^4`

Putting `a=sqrt(5)x` and `b=sqrt(2)y` we get:

`(5x^2-ksqrt(10)xy+2y^2)(5x^2+ksqrt(10)xy+2y^2)`

`=25x^4+10(2-k^2)x^2y^2+4y^4`

Solve:

`16 = 10(2-k^2) = 20-10k^2`

So:

`10k^2 = 20-16=4`

`k^2 = 2/5 = 10/25`

`k=+-sqrt(10)/5`

So:

`ksqrt(10) = +-sqrt(10)/5*sqrt(10) = +-2`

Hence:

`25x^4+16x^2y^2+4y^4 = (5x^2-2xy+2y^2)(5x^2+2xy+2y^2)`

Answer 2

`(5x^2+2xy+2y^2)(5x^2-2xy+2y^2)`

Explanation:

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

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

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

We can write the given expression as:

`25x^4+16x^2y^2+4y^4=25x^4+(20x^2y^2-4x^2y^2)+4y^4`

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

`25x^4+(20x^2y^2-4x^2y^2)+4y^4=(25x^4+20x^2y^2+4y^4)-4x^2y^2`

Recall that `25x^4+20x^2y^2+4y^4=(5x^2+2y^2)^2`. Also, it will be important that `4x^2y^2=(2xy)^2`.

`(25x^4+20x^2y^2+4y^4)-4x^2y^2=(5x^2+2y^2)^2-(2xy)^2`

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

`(5x^2+2y^2)^2-(2xy)^2=(5x^2+2xy+2y^2)(5x^2-2xy+2y^2)`