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 negative, $a$ and $b$ are both negative. List all such integer pairs that give product $36$.
Rewrite $6x^{2}-13x+6$ as $\left(6x^{2}-9x\right)+\left(-4x+6\right)$.
$$\left(6x^{2}-9x\right)+\left(-4x+6\right)$$
Factor out $3x$ in the first and $-2$ in the second group.
$$3x\left(2x-3\right)-2\left(2x-3\right)$$
Factor out common term $2x-3$ by using distributive property.
$$\left(2x-3\right)\left(3x-2\right)$$
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 $ax^{2}+bx+c=0$.
$$6x^{2}-13x+6=0$$
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 $5$.
$$x=\frac{18}{12}$$
Reduce the fraction $\frac{18}{12}$ to lowest terms by extracting and canceling out $6$.
$$x=\frac{3}{2}$$
Now solve the equation $x=\frac{13±5}{12}$ when $±$ is minus. Subtract $5$ from $13$.
$$x=\frac{8}{12}$$
Reduce the fraction $\frac{8}{12}$ to lowest terms by extracting and canceling out $4$.
$$x=\frac{2}{3}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{3}{2}$ for $x_{1}$ and $\frac{2}{3}$ for $x_{2}$.
Multiply $\frac{2x-3}{2}$ times $\frac{3x-2}{3}$ by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form $x^2+Bx+C=0$.This is achieved by dividing both sides of the equation by $6$
$$x ^ 2 -\frac{13}{6}x +1 = 0$$
Let $r$ and $s$ be the factors for the quadratic equation such that $x^2+Bx+C=(x−r)(x−s)$ where sum of factors $(r+s)=−B$ and the product of factors $rs = C$
$$r + s = \frac{13}{6} $$ $$ rs = 1$$
Two numbers $r$ and $s$ sum up to $\frac{13}{6}$ exactly when the average of the two numbers is $\frac{1}{2}*\frac{13}{6} = \frac{13}{12}$. You can also see that the midpoint of $r$ and $s$ corresponds to the axis of symmetry of the parabola represented by the quadratic equation $y=x^2+Bx+C$. The values of $r$ and $s$ are equidistant from the center by an unknown quantity $u$. Express $r$ and $s$ with respect to variable $u$.
