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.
$$a+b=-1$$ $$ab=1\left(-6\right)=-6$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-6$.
$$1,-6$$ $$2,-3$$
Calculate the sum for each pair.
$$1-6=-5$$ $$2-3=-1$$
The solution is the pair that gives sum $-1$.
$$a=-3$$ $$b=2$$
Rewrite $y^{2}-y-6$ as $\left(y^{2}-3y\right)+\left(2y-6\right)$.
$$\left(y^{2}-3y\right)+\left(2y-6\right)$$
Factor out $y$ in the first and $2$ in the second group.
$$y\left(y-3\right)+2\left(y-3\right)$$
Factor out common term $y-3$ by using distributive property.
$$\left(y-3\right)\left(y+2\right)$$
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 $ax^{2}+bx+c=0$.
$$y^{2}-y-6=0$$
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.
