Factor the expression by grouping. First, the expression needs to be rewritten as $12x^{2}+ax+bx+6$. To find $a$ and $b$, set up a system to be solved.
$$a+b=17$$ $$ab=12\times 6=72$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $72$.
Rewrite $12x^{2}+17x+6$ as $\left(12x^{2}+8x\right)+\left(9x+6\right)$.
$$\left(12x^{2}+8x\right)+\left(9x+6\right)$$
Factor out $4x$ in the first and $3$ in the second group.
$$4x\left(3x+2\right)+3\left(3x+2\right)$$
Factor out common term $3x+2$ by using distributive property.
$$\left(3x+2\right)\left(4x+3\right)$$
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 $ax^{2}+bx+c=0$.
$$12x^{2}+17x+6=0$$
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 $x=\frac{-17±1}{24}$ when $±$ is plus. Add $-17$ to $1$.
$$x=-\frac{16}{24}$$
Reduce the fraction $\frac{-16}{24}$ to lowest terms by extracting and canceling out $8$.
$$x=-\frac{2}{3}$$
Now solve the equation $x=\frac{-17±1}{24}$ when $±$ is minus. Subtract $1$ from $-17$.
$$x=-\frac{18}{24}$$
Reduce the fraction $\frac{-18}{24}$ to lowest terms by extracting and canceling out $6$.
$$x=-\frac{3}{4}$$
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 $-\frac{3}{4}$ for $x_{2}$.
Multiply $\frac{3x+2}{3}$ times $\frac{4x+3}{4}$ by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
