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

2 Answers
May 10, 2016

Answer:

#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)#

May 10, 2016

Answer:

#(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)#