How do you factor #36x^2+96xy+64y^2#?

2 Answers
Cesareo R.
Aug 29, 2016

#4(3x+4y)^2#

Explanation:

#36x^2+96xy+64y^2 = 4(9x^2+24 x y + 16 y^2) = 4(3x+4y)^2#

George C.
Aug 29, 2016

#36x^2+96xy+64y^2=4(3x+4y)^2#

Explanation:

Note that all of the coefficients are divisible by #4#, so we can separate that out as a scalar factor. Then noticing that the first and last terms are perfect squares, we find that the middle term matches the pattern:

#a^2+2ab+b^2 = (a+b)^2#

with #a=3x# and #b=4y# as follows:

#36x^2+96xy+64y^2#

#=4(9x^2+24xy+16y^2)#

#=4((3x)^2+2(3x)(4y)+(4y)^2)#

#=4(3x+4y)^2#

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