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

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Answer 1 · 2016-10-07T06:16:52

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

Notice that both #x^2# and #9y^2 = (3y)^2# are perfect squares.

So we can try squaring #(x+3y)# and see if the middle term works out as #6xy#...

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

So #x^2+6xy+9x^2# is a perfect square trinomial.

In general we have:

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

So we can recognise perfect square trinomials by whether the middle term is twice the product of the square roots of the first and last terms.