How do you factor completely $25a^4-49b^2$ ?

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Answer 1 · 2017-04-19T05:59:40

$25a^4-49b^2 = (5a^2-7b)(5a^2+7b)$

Note that both $25a^4 = (5a^2)^2$ and $49b^2=(7b)^2$ are perfect squares.

The difference of squares identity can be written:

$A^2-B^2=(A-B)(A+B)$

Using this with $A=5a^2$ and $B=7b$ we find:

$25a^4-49b^2 = (5a^2)^2-(7b)^2$

$color(white)(25a^4-49b^2) = (5a^2-7b)(5a^2+7b)$

There are no simpler factors.

How do you factor completely 25a^4-49b^225a^4-49b^2?