Consider $x^{4}+4x^{3}-5x^{2}$. Factor out $x^{2}$.
$$x^{2}\left(x^{2}+4x-5\right)$$
Consider $x^{2}+4x-5$. Factor the expression by grouping. First, the expression needs to be rewritten as $x^{2}+ax+bx-5$. To find $a$ and $b$, set up a system to be solved.
$$a+b=4$$ $$ab=1\left(-5\right)=-5$$
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. The only such pair is the system solution.
$$a=-1$$ $$b=5$$
Rewrite $x^{2}+4x-5$ as $\left(x^{2}-x\right)+\left(5x-5\right)$.
$$\left(x^{2}-x\right)+\left(5x-5\right)$$
Factor out $x$ in the first and $5$ in the second group.
$$x\left(x-1\right)+5\left(x-1\right)$$
Factor out common term $x-1$ by using distributive property.
