Factor (x+y)^2+6(x+y)+9

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

Step 1

Let $u = x + y$ . Substitute $u$ for all occurrences of $x + y$ .

$u^{2} + 6 u + 9$

Step 2

Factor using the perfect square rule.

Rewrite $9$ as $3^{2}$ .

$u^{2} + 6 u + 3^{2}$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 u = 2 \cdot u \cdot 3$

Rewrite the polynomial.

$u^{2} + 2 \cdot u \cdot 3 + 3^{2}$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = u$ and $b = 3$ .

${(u + 3)}^{2}$

Step 3

Replace all occurrences of $u$ with $x + y$ .

${(x + y + 3)}^{2}$