How do you factor completely: #8x^4-28x^6#?

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Answer 1 · 2015-07-28T23:59:11

#8x^4-28x^6 = 4x^4(2-7x^2) = 4x^4(sqrt(2)-sqrt(7)x)(sqrt(2)+sqrt(7)x)#

Both terms are divisible by #4x^4# so we can separate out that factor first...

#8x^4-28x^6 = 4x^4(2-7x^2)#

Next, if we allow irrational coefficients, #2-7x^2# can be treated as a difference of squares, so we can use the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a=sqrt(2)# and #b=sqrt(7)x#

So:

#2-7x^2 = (sqrt(2))^2 - (sqrt(7)x)^2 = (sqrt(2)-sqrt(7)x)(sqrt(2)+sqrt(7)x)#

Putting it all together:

#8x^4-28x^6 = 4x^4(2-7x^2) = 4x^4(sqrt(2)-sqrt(7)x)(sqrt(2)+sqrt(7)x)#