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)#