How do you factor $a^{4} + 2 a^{2} b^{2} + b^{8}$ $a^4+2a^2b^2+b^8$ ?

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How do you factor $a^{4} + 2 a^{2} b^{2} + b^{8}$ $a^4+2a^2b^2+b^8$ ?

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Answer 1 · 2017-04-03T13:24:33

That expression is prime.

Explanation:

If you are certain about the exponent on b in the middle term, that expression is prime. That is to say, it does not factor.

However, $a^{4} + 2 a^{2} b^{4} + b^{8}$ $a^4+2a^2b^4+b^8$ would be a Perfect Square Trinomial.
For $a^{4} + 2 a^{2} b^{4} + b^{8}$ $a^4+2a^2b^4+b^8$ , we would observe that
the positive (principal) square root of
$a^{4}$ $a^4$ is $a^{2}$ $a^2$ , and the square root of $b^{8}$ $b^8$ is $b^{4}$ $b^4$ .
The middle term is twice the product of $a^{2}$ $a^2$ and $b^{4}$ $b^4$ .
Therefore we would have a Perfect Square Trinomial factoring as
${(a^{2} + b^{4})}^{2}$ $(a^2+b^4)^2$ .

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How do you factor $a^{4} + 2 a^{2} b^{2} + b^{8}$ $a^4+2a^2b^2+b^8$ ?

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Explanation:

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How do you factor $a^{4} + 2 a^{2} b^{2} + b^{8}$ $a^4+2a^2b^2+b^8$ ?