How do you factor $9b^4-16c^4$ ?

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Answer 1 · 2015-12-24T20:24:46

Recognize that this is a difference of squares, which factor as follows:

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

In your scenario:

$9b^4-16c^4$

$=>(3b^2)^2-(4c^2)^2=(3b^2+4c^2)(3b^2-4c^2)$

Note that this is a fine final answer, but that $3b^2-4c^2$ can also be treated as a difference of squares.

$3b^2-4c^2$

$=>(bsqrt3)^2-(2c)^2=(bsqrt3+2c)(bsqrt3-2c)$

So,

$9b^4-16c^4=(3b^2+4c^2)(bsqrt3+2c)(bsqrt3-2c)$

or

$9b^4-16c^4=(3b^2+4c^2)(3b^2-4c^2)$

How do you factor 9b^4-16c^49b^4-16c^4?