Factor the expression by grouping. First, the expression needs to be rewritten as $6x^{2}+ax+bx+6$. To find $a$ and $b$, set up a system to be solved.
$$a+b=13$$ $$ab=6\times 6=36$$
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 $36$.
Rewrite $6x^{2}+13x+6$ as $\left(6x^{2}+4x\right)+\left(9x+6\right)$.
$$\left(6x^{2}+4x\right)+\left(9x+6\right)$$
Factor out $2x$ in the first and $3$ in the second group.
$$2x\left(3x+2\right)+3\left(3x+2\right)$$
Factor out common term $3x+2$ by using distributive property.
$$\left(3x+2\right)\left(2x+3\right)$$
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 $ax^{2}+bx+c=0$.
$$6x^{2}+13x+6=0$$
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 $x=\frac{-13±5}{12}$ when $±$ is plus. Add $-13$ to $5$.
$$x=-\frac{8}{12}$$
Reduce the fraction $\frac{-8}{12}$ to lowest terms by extracting and canceling out $4$.
$$x=-\frac{2}{3}$$
Now solve the equation $x=\frac{-13±5}{12}$ when $±$ is minus. Subtract $5$ from $-13$.
$$x=-\frac{18}{12}$$
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 $-\frac{2}{3}$ for $x_{1}$ and $-\frac{3}{2}$ for $x_{2}$.
Multiply $\frac{3x+2}{3}$ times $\frac{2x+3}{2}$ by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
