Factor the expression by grouping. First, the expression needs to be rewritten as $y^{2}+ay+by-160$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-36$$ $$ab=1\left(-160\right)=-160$$
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 $-160$.
Rewrite $y^{2}-36y-160$ as $\left(y^{2}-40y\right)+\left(4y-160\right)$.
$$\left(y^{2}-40y\right)+\left(4y-160\right)$$
Factor out $y$ in the first and $4$ in the second group.
$$y\left(y-40\right)+4\left(y-40\right)$$
Factor out common term $y-40$ by using distributive property.
$$\left(y-40\right)\left(y+4\right)$$
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 $ax^{2}+bx+c=0$.
$$y^{2}-36y-160=0$$
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.
