Factor the expression by grouping. First, the expression needs to be rewritten as $-8x^{2}+ax+bx+5$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-3$$ $$ab=-8\times 5=-40$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-40$.
$$1,-40$$ $$2,-20$$ $$4,-10$$ $$5,-8$$
Calculate the sum for each pair.
$$1-40=-39$$ $$2-20=-18$$ $$4-10=-6$$ $$5-8=-3$$
The solution is the pair that gives sum $-3$.
$$a=5$$ $$b=-8$$
Rewrite $-8x^{2}-3x+5$ as $\left(-8x^{2}+5x\right)+\left(-8x+5\right)$.
$$\left(-8x^{2}+5x\right)+\left(-8x+5\right)$$
Factor out $-x$ in the first and $-1$ in the second group.
$$-x\left(8x-5\right)-\left(8x-5\right)$$
Factor out common term $8x-5$ by using distributive property.
