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:

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

How do you factor completely: 8x^4-28x^68x^4-28x^6?