Factor the expression by grouping. First, the expression needs to be rewritten as $x^{2}+ax+bx+9$. To find $a$ and $b$, set up a system to be solved.
$$a+b=6$$ $$ab=1\times 9=9$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $9$.
$$1,9$$ $$3,3$$
Calculate the sum for each pair.
$$1+9=10$$ $$3+3=6$$
The solution is the pair that gives sum $6$.
$$a=3$$ $$b=3$$
Rewrite $x^{2}+6x+9$ as $\left(x^{2}+3x\right)+\left(3x+9\right)$.
$$\left(x^{2}+3x\right)+\left(3x+9\right)$$
Factor out $x$ in the first and $3$ in the second group.
$$x\left(x+3\right)+3\left(x+3\right)$$
Factor out common term $x+3$ by using distributive property.
