Factor the expression by grouping. First, the expression needs to be rewritten as $12x^{2}+ax+bx-3$. To find $a$ and $b$, set up a system to be solved.
$$a+b=16$$ $$ab=12\left(-3\right)=-36$$
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 $-36$.
Rewrite $12x^{2}+16x-3$ as $\left(12x^{2}-2x\right)+\left(18x-3\right)$.
$$\left(12x^{2}-2x\right)+\left(18x-3\right)$$
Factor out $2x$ in the first and $3$ in the second group.
$$2x\left(6x-1\right)+3\left(6x-1\right)$$
Factor out common term $6x-1$ by using distributive property.
$$\left(6x-1\right)\left(2x+3\right)$$
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 $ax^{2}+bx+c=0$.
$$12x^{2}+16x-3=0$$
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 $x=\frac{-16±20}{24}$ when $±$ is plus. Add $-16$ to $20$.
$$x=\frac{4}{24}$$
Reduce the fraction $\frac{4}{24}$ to lowest terms by extracting and canceling out $4$.
$$x=\frac{1}{6}$$
Now solve the equation $x=\frac{-16±20}{24}$ when $±$ is minus. Subtract $20$ from $-16$.
$$x=-\frac{36}{24}$$
Reduce the fraction $\frac{-36}{24}$ to lowest terms by extracting and canceling out $12$.
$$x=-\frac{3}{2}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{1}{6}$ for $x_{1}$ and $-\frac{3}{2}$ for $x_{2}$.
Multiply $\frac{6x-1}{6}$ times $\frac{2x+3}{2}$ by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
