Factor the expression by grouping. First, the expression needs to be rewritten as $y^{2}+ay+by+6$. To find $a$ and $b$, set up a system to be solved.
$$a+b=5$$ $$ab=1\times 6=6$$
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 $6$.
$$1,6$$ $$2,3$$
Calculate the sum for each pair.
$$1+6=7$$ $$2+3=5$$
The solution is the pair that gives sum $5$.
$$a=2$$ $$b=3$$
Rewrite $y^{2}+5y+6$ as $\left(y^{2}+2y\right)+\left(3y+6\right)$.
$$\left(y^{2}+2y\right)+\left(3y+6\right)$$
Factor out $y$ in the first and $3$ in the second group.
$$y\left(y+2\right)+3\left(y+2\right)$$
Factor out common term $y+2$ by using distributive property.
$$\left(y+2\right)\left(y+3\right)$$
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 $ax^{2}+bx+c=0$.
$$y^{2}+5y+6=0$$
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.
$$y=\frac{-5±\sqrt{5^{2}-4\times 6}}{2}$$
Square $5$.
$$y=\frac{-5±\sqrt{25-4\times 6}}{2}$$
Multiply $-4$ times $6$.
$$y=\frac{-5±\sqrt{25-24}}{2}$$
Add $25$ to $-24$.
$$y=\frac{-5±\sqrt{1}}{2}$$
Take the square root of $1$.
$$y=\frac{-5±1}{2}$$
Now solve the equation $y=\frac{-5±1}{2}$ when $±$ is plus. Add $-5$ to $1$.
$$y=-\frac{4}{2}$$
Divide $-4$ by $2$.
$$y=-2$$
Now solve the equation $y=\frac{-5±1}{2}$ when $±$ is minus. Subtract $1$ from $-5$.
$$y=-\frac{6}{2}$$
Divide $-6$ by $2$.
$$y=-3$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $-2$ for $x_{1}$ and $-3$ for $x_{2}$.
