Consider $c^{2}+3c-10$. Factor the expression by grouping. First, the expression needs to be rewritten as $c^{2}+pc+qc-10$. To find $p$ and $q$, set up a system to be solved.
$$p+q=3$$ $$pq=1\left(-10\right)=-10$$
Since $pq$ is negative, $p$ and $q$ have the opposite signs. Since $p+q$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-10$.
$$-1,10$$ $$-2,5$$
Calculate the sum for each pair.
$$-1+10=9$$ $$-2+5=3$$
The solution is the pair that gives sum $3$.
$$p=-2$$ $$q=5$$
Rewrite $c^{2}+3c-10$ as $\left(c^{2}-2c\right)+\left(5c-10\right)$.
$$\left(c^{2}-2c\right)+\left(5c-10\right)$$
Factor out $c$ in the first and $5$ in the second group.
$$c\left(c-2\right)+5\left(c-2\right)$$
Factor out common term $c-2$ by using distributive property.
