Factor the expression by grouping. First, the expression needs to be rewritten as $12y^{2}+ay+by+5$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-16$$ $$ab=12\times 5=60$$
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 $60$.
Rewrite $12y^{2}-16y+5$ as $\left(12y^{2}-10y\right)+\left(-6y+5\right)$.
$$\left(12y^{2}-10y\right)+\left(-6y+5\right)$$
Factor out $2y$ in the first and $-1$ in the second group.
$$2y\left(6y-5\right)-\left(6y-5\right)$$
Factor out common term $6y-5$ by using distributive property.
$$\left(6y-5\right)\left(2y-1\right)$$
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 $ax^{2}+bx+c=0$.
$$12y^{2}-16y+5=0$$
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 $y=\frac{16±4}{24}$ when $±$ is plus. Add $16$ to $4$.
$$y=\frac{20}{24}$$
Reduce the fraction $\frac{20}{24}$ to lowest terms by extracting and canceling out $4$.
$$y=\frac{5}{6}$$
Now solve the equation $y=\frac{16±4}{24}$ when $±$ is minus. Subtract $4$ from $16$.
$$y=\frac{12}{24}$$
Reduce the fraction $\frac{12}{24}$ to lowest terms by extracting and canceling out $12$.
$$y=\frac{1}{2}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{5}{6}$ for $x_{1}$ and $\frac{1}{2}$ for $x_{2}$.
Multiply $\frac{6y-5}{6}$ times $\frac{2y-1}{2}$ 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 $12$
$$x ^ 2 -\frac{4}{3}x +\frac{5}{12} = 0$$
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 $rs = C$
$$r + s = \frac{4}{3} $$ $$ rs = \frac{5}{12}$$
Two numbers $r$ and $s$ sum up to $\frac{4}{3}$ exactly when the average of the two numbers is $\frac{1}{2}*\frac{4}{3} = \frac{2}{3}$. 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 $u$.
$$r = \frac{2}{3} - u$$ $$s = \frac{2}{3} + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = \frac{5}{12}$
