Factor the expression by grouping. First, the expression needs to be rewritten as $x^{2}+ax+bx+18$. To find $a$ and $b$, set up a system to be solved.
$$a+b=9$$ $$ab=1\times 18=18$$
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 $18$.
$$1,18$$ $$2,9$$ $$3,6$$
Calculate the sum for each pair.
$$1+18=19$$ $$2+9=11$$ $$3+6=9$$
The solution is the pair that gives sum $9$.
$$a=3$$ $$b=6$$
Rewrite $x^{2}+9x+18$ as $\left(x^{2}+3x\right)+\left(6x+18\right)$.
$$\left(x^{2}+3x\right)+\left(6x+18\right)$$
Factor out $x$ in the first and $6$ in the second group.
$$x\left(x+3\right)+6\left(x+3\right)$$
Factor out common term $x+3$ by using distributive property.
$$\left(x+3\right)\left(x+6\right)$$
Steps Using the Quadratic Formula
Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$.
$$x^{2}+9x+18=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
$$x=\frac{-9±\sqrt{9^{2}-4\times 18}}{2}$$
Square $9$.
$$x=\frac{-9±\sqrt{81-4\times 18}}{2}$$
Multiply $-4$ times $18$.
$$x=\frac{-9±\sqrt{81-72}}{2}$$
Add $81$ to $-72$.
$$x=\frac{-9±\sqrt{9}}{2}$$
Take the square root of $9$.
$$x=\frac{-9±3}{2}$$
Now solve the equation $x=\frac{-9±3}{2}$ when $±$ is plus. Add $-9$ to $3$.
$$x=-\frac{6}{2}$$
Divide $-6$ by $2$.
$$x=-3$$
Now solve the equation $x=\frac{-9±3}{2}$ when $±$ is minus. Subtract $3$ from $-9$.
$$x=-\frac{12}{2}$$
Divide $-12$ by $2$.
$$x=-6$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $-3$ for $x_{1}$ and $-6$ for $x_{2}$.
