Consider $y^{2}-8y+15$. Factor the expression by grouping. First, the expression needs to be rewritten as $y^{2}+ay+by+15$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-8$$ $$ab=1\times 15=15$$
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 $15$.
$$-1,-15$$ $$-3,-5$$
Calculate the sum for each pair.
$$-1-15=-16$$ $$-3-5=-8$$
The solution is the pair that gives sum $-8$.
$$a=-5$$ $$b=-3$$
Rewrite $y^{2}-8y+15$ as $\left(y^{2}-5y\right)+\left(-3y+15\right)$.
$$\left(y^{2}-5y\right)+\left(-3y+15\right)$$
Factor out $y$ in the first and $-3$ in the second group.
$$y\left(y-5\right)-3\left(y-5\right)$$
Factor out common term $y-5$ by using distributive property.
