Consider $x^{2}+2x-15$. Factor the expression by grouping. First, the expression needs to be rewritten as $x^{2}+ax+bx-15$. To find $a$ and $b$, set up a system to be solved.
$$a+b=2$$ $$ab=1\left(-15\right)=-15$$
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. List all such integer pairs that give product $-15$.
$$-1,15$$ $$-3,5$$
Calculate the sum for each pair.
$$-1+15=14$$ $$-3+5=2$$
The solution is the pair that gives sum $2$.
$$a=-3$$ $$b=5$$
Rewrite $x^{2}+2x-15$ as $\left(x^{2}-3x\right)+\left(5x-15\right)$.
$$\left(x^{2}-3x\right)+\left(5x-15\right)$$
Factor out $x$ in the first and $5$ in the second group.
$$x\left(x-3\right)+5\left(x-3\right)$$
Factor out common term $x-3$ by using distributive property.
