Consider $-x^{4}-7x^{3}+8x^{2}$. Factor out $x^{2}$.
$$x^{2}\left(-x^{2}-7x+8\right)$$
Consider $-x^{2}-7x+8$. Factor the expression by grouping. First, the expression needs to be rewritten as $-x^{2}+ax+bx+8$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-7$$ $$ab=-8=-8$$
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 $-8$.
$$1,-8$$ $$2,-4$$
Calculate the sum for each pair.
$$1-8=-7$$ $$2-4=-2$$
The solution is the pair that gives sum $-7$.
$$a=1$$ $$b=-8$$
Rewrite $-x^{2}-7x+8$ as $\left(-x^{2}+x\right)+\left(-8x+8\right)$.
$$\left(-x^{2}+x\right)+\left(-8x+8\right)$$
Factor out $x$ in the first and $8$ in the second group.
$$x\left(-x+1\right)+8\left(-x+1\right)$$
Factor out common term $-x+1$ by using distributive property.
