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

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Answer 1 · 2016-10-07T06:16:52

$x^{2} + 6 x y + 9 y^{2} = {(x + 3 y)}^{2}$ $x^2+6xy+9y^2 = (x+3y)^2$

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

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

${(x + 3 y)}^{2} = (x + 3 y) (x + 3 y) = x^{2} + 6 x y + 9 y^{2}$ $(x+3y)^2 = (x+3y)(x+3y) = x^2+6xy+9y^2$

So $x^{2} + 6 x y + 9 x^{2}$ $x^2+6xy+9x^2$ is a perfect square trinomial.

In general we have:

${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $(a+b)^2 = a^2+2ab+b^2$

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 x2+6xy+9y2x^2+6xy+9y^2 ?