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.
$$a+b=3$$ $$ab=1\left(-18\right)=-18$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-18$.
$$-1,18$$ $$-2,9$$ $$-3,6$$
Calculate the sum for each pair.
$$-1+18=17$$ $$-2+9=7$$ $$-3+6=3$$
The solution is the pair that gives sum $3$.
$$a=-3$$ $$b=6$$
Rewrite $x^{2}+3x-18$ 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 $ax^{2}+bx+c=0$.
$$x^{2}+3x-18=0$$
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.
$$x=\frac{-3±\sqrt{3^{2}-4\left(-18\right)}}{2}$$
Square $3$.
$$x=\frac{-3±\sqrt{9-4\left(-18\right)}}{2}$$
Multiply $-4$ times $-18$.
$$x=\frac{-3±\sqrt{9+72}}{2}$$
Add $9$ to $72$.
$$x=\frac{-3±\sqrt{81}}{2}$$
Take the square root of $81$.
$$x=\frac{-3±9}{2}$$
Now solve the equation $x=\frac{-3±9}{2}$ when $±$ is plus. Add $-3$ to $9$.
$$x=\frac{6}{2}$$
Divide $6$ by $2$.
$$x=3$$
Now solve the equation $x=\frac{-3±9}{2}$ when $±$ is minus. Subtract $9$ from $-3$.
$$x=-\frac{12}{2}$$
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 $x_{2}$.
