Consider $C^{2}+3C-10$. Factor the expression by grouping. First, the expression needs to be rewritten as $C^{2}+aC+bC-10$. To find $a$ and $b$, set up a system to be solved.
$$a+b=3$$ $$ab=1\left(-10\right)=-10$$
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 $-10$.
$$-1,10$$ $$-2,5$$
Calculate the sum for each pair.
$$-1+10=9$$ $$-2+5=3$$
The solution is the pair that gives sum $3$.
$$a=-2$$ $$b=5$$
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.
