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.

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