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.
$$a+b=7$$ $$ab=1\left(-8\right)=-8$$
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. 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.
$$\left(x-1\right)\left(x+8\right)$$
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 $ax^{2}+bx+c=0$.
$$x^{2}+7x-8=0$$
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.
$$x=\frac{-7±\sqrt{7^{2}-4\left(-8\right)}}{2}$$
Square $7$.
$$x=\frac{-7±\sqrt{49-4\left(-8\right)}}{2}$$
Multiply $-4$ times $-8$.
$$x=\frac{-7±\sqrt{49+32}}{2}$$
Add $49$ to $32$.
$$x=\frac{-7±\sqrt{81}}{2}$$
Take the square root of $81$.
$$x=\frac{-7±9}{2}$$
Now solve the equation $x=\frac{-7±9}{2}$ when $±$ is plus. Add $-7$ to $9$.
$$x=\frac{2}{2}$$
Divide $2$ by $2$.
$$x=1$$
Now solve the equation $x=\frac{-7±9}{2}$ when $±$ is minus. Subtract $9$ from $-7$.
$$x=-\frac{16}{2}$$
Divide $-16$ by $2$.
$$x=-8$$
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 $-8$ for $x_{2}$.
