Factor the expression by grouping. First, the expression needs to be rewritten as $3x^{2}+ax+bx-12$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-16$$ $$ab=3\left(-12\right)=-36$$
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 $3x^{2}-16x-12$ as $\left(3x^{2}-18x\right)+\left(2x-12\right)$.
$$\left(3x^{2}-18x\right)+\left(2x-12\right)$$
Factor out $3x$ in the first and $2$ in the second group.
$$3x\left(x-6\right)+2\left(x-6\right)$$
Factor out common term $x-6$ by using distributive property.
$$\left(x-6\right)\left(3x+2\right)$$
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 $ax^{2}+bx+c=0$.
$$3x^{2}-16x-12=0$$
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 $x=\frac{16±20}{6}$ when $±$ is plus. Add $16$ to $20$.
$$x=\frac{36}{6}$$
Divide $36$ by $6$.
$$x=6$$
Now solve the equation $x=\frac{16±20}{6}$ when $±$ is minus. Subtract $20$ from $16$.
$$x=-\frac{4}{6}$$
Reduce the fraction $\frac{-4}{6}$ to lowest terms by extracting and canceling out $2$.
$$x=-\frac{2}{3}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $6$ for $x_{1}$ and $-\frac{2}{3}$ for $x_{2}$.
