Factor the expression by grouping. First, the expression needs to be rewritten as $18x^{2}+ax+bx-2$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-9$$ $$ab=18\left(-2\right)=-36$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-36$.
Rewrite $18x^{2}-9x-2$ as $\left(18x^{2}-12x\right)+\left(3x-2\right)$.
$$\left(18x^{2}-12x\right)+\left(3x-2\right)$$
Factor out $6x$ in $18x^{2}-12x$.
$$6x\left(3x-2\right)+3x-2$$
Factor out common term $3x-2$ by using distributive property.
$$\left(3x-2\right)\left(6x+1\right)$$
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 $ax^{2}+bx+c=0$.
$$18x^{2}-9x-2=0$$
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.
Now solve the equation $x=\frac{9±15}{36}$ when $±$ is plus. Add $9$ to $15$.
$$x=\frac{24}{36}$$
Reduce the fraction $\frac{24}{36}$ to lowest terms by extracting and canceling out $12$.
$$x=\frac{2}{3}$$
Now solve the equation $x=\frac{9±15}{36}$ when $±$ is minus. Subtract $15$ from $9$.
$$x=-\frac{6}{36}$$
Reduce the fraction $\frac{-6}{36}$ to lowest terms by extracting and canceling out $6$.
$$x=-\frac{1}{6}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{2}{3}$ for $x_{1}$ and $-\frac{1}{6}$ for $x_{2}$.
Multiply $\frac{3x-2}{3}$ times $\frac{6x+1}{6}$ by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
