How do you factor #x^2+6xy+9y^2-25#?

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Answer 1 · 2018-03-28T06:15:51

#x^2+6xy+9y^2-25 = (x+3y-5)(x+3y+5)#

Given:

#x^2+6xy+9y^2-25#

Note that both #x^2+6xy+9y^2 = (x+3y)^2# and #25 = 5^2# are perfect squares.

So the given polynomial will factor as a difference of squares:

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

with #A=x+3y# and #B=5# ...

#x^2+6xy+9y^2-25 = (x+3y)^2-5^2#

#color(white)(x^2+6xy+9y^2-25) = ((x+3y)-5)((x+3y)+5)#

#color(white)(x^2+6xy+9y^2-25) = (x+3y-5)(x+3y+5)#