How do you factor the expression #16x² + 48xy +36y²#?

1 Answer
Jan 25, 2017

#16x^2+48xy+36y^2 = 4(2x+3)^2#

Explanation:

First note that all of the terms are divisible by #4#, so separate that out as a factor:

#16x^2+48xy+36y^2 = 4(4x^2+12xy+9y^2)#

Next note that both #4x^2 = (2x)^2# and #9y^2 = (3y)^2# are perfect squares. So if we square #(2x+3)# then the resulting first term will be #4x^2# and the last term #9y^2#. How about the middle term?

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

#color(white)((2x+3)^2) = 4x^2+12xy+9y^2#

So we have a perfect square trinomial.

Putting it all together:

#16x^2+48xy+36y^2 = 4(2x+3)^2#