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.
$$\left(x-1\right)\left(x+5\right)$$
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.
$$x=\frac{-4±\sqrt{4^{2}-4\left(-5\right)}}{2}$$
Square $4$.
$$x=\frac{-4±\sqrt{16-4\left(-5\right)}}{2}$$
Multiply $-4$ times $-5$.
$$x=\frac{-4±\sqrt{16+20}}{2}$$
Add $16$ to $20$.
$$x=\frac{-4±\sqrt{36}}{2}$$
Take the square root of $36$.
$$x=\frac{-4±6}{2}$$
Now solve the equation $x=\frac{-4±6}{2}$ when $±$ is plus. Add $-4$ to $6$.
$$x=\frac{2}{2}$$
Divide $2$ by $2$.
$$x=1$$
Now solve the equation $x=\frac{-4±6}{2}$ when $±$ is minus. Subtract $6$ from $-4$.
$$x=-\frac{10}{2}$$
Divide $-10$ by $2$.
$$x=-5$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $1$ for $x_{1}$ and $-5$ for $x_{2}$.
