Consider $y^{2}+2y-3$. Factor the expression by grouping. First, the expression needs to be rewritten as $y^{2}+ay+by-3$. To find $a$ and $b$, set up a system to be solved.
$$a+b=2$$ $$ab=1\left(-3\right)=-3$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.
$$a=-1$$ $$b=3$$
Rewrite $y^{2}+2y-3$ as $\left(y^{2}-y\right)+\left(3y-3\right)$.
$$\left(y^{2}-y\right)+\left(3y-3\right)$$
Factor out $y$ in the first and $3$ in the second group.
$$y\left(y-1\right)+3\left(y-1\right)$$
Factor out common term $y-1$ by using distributive property.
