Factor the expression by grouping. First, the expression needs to be rewritten as $6x^{2}+ax+bx-12$. To find $a$ and $b$, set up a system to be solved.
$$a+b=1$$ $$ab=6\left(-12\right)=-72$$
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 $-72$.
Rewrite $6x^{2}+x-12$ as $\left(6x^{2}-8x\right)+\left(9x-12\right)$.
$$\left(6x^{2}-8x\right)+\left(9x-12\right)$$
Factor out $2x$ in the first and $3$ in the second group.
$$2x\left(3x-4\right)+3\left(3x-4\right)$$
Factor out common term $3x-4$ by using distributive property.
$$\left(3x-4\right)\left(2x+3\right)$$
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 $ax^{2}+bx+c=0$.
$$6x^{2}+x-12=0$$
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 $x=\frac{-1±17}{12}$ when $±$ is plus. Add $-1$ to $17$.
$$x=\frac{16}{12}$$
Reduce the fraction $\frac{16}{12}$ to lowest terms by extracting and canceling out $4$.
$$x=\frac{4}{3}$$
Now solve the equation $x=\frac{-1±17}{12}$ when $±$ is minus. Subtract $17$ from $-1$.
$$x=-\frac{18}{12}$$
Reduce the fraction $\frac{-18}{12}$ to lowest terms by extracting and canceling out $6$.
$$x=-\frac{3}{2}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{4}{3}$ for $x_{1}$ and $-\frac{3}{2}$ for $x_{2}$.
Multiply $\frac{3x-4}{3}$ times $\frac{2x+3}{2}$ by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
