Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
$$-x^{2}-4x+5$$
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.
$$a+b=-4$$ $$ab=-5=-5$$
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. 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.
$$\left(-x+1\right)\left(x+5\right)$$
Steps Using the Quadratic Formula
Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$.
$$-x^{2}-4x+5=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
