To solve the equation, factor $x^{2}-2x-8$ using formula $x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right)$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-2$$ $$ab=-8$$
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 $-2$.
$$a=-4$$ $$b=2$$
Rewrite factored expression $\left(x+a\right)\left(x+b\right)$ using the obtained values.
$$\left(x-4\right)\left(x+2\right)$$
To find equation solutions, solve $x-4=0$ and $x+2=0$.
$$x=4$$ $$x=-2$$
Steps Using Factoring By Grouping
Combine $2x$ and $-4x$ to get $-2x$.
$$x^{2}-2x-8=0$$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx-8$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-2$$ $$ab=1\left(-8\right)=-8$$
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 $-2$.
$$a=-4$$ $$b=2$$
Rewrite $x^{2}-2x-8$ as $\left(x^{2}-4x\right)+\left(2x-8\right)$.
$$\left(x^{2}-4x\right)+\left(2x-8\right)$$
Factor out $x$ in the first and $2$ in the second group.
$$x\left(x-4\right)+2\left(x-4\right)$$
Factor out common term $x-4$ by using distributive property.
$$\left(x-4\right)\left(x+2\right)$$
To find equation solutions, solve $x-4=0$ and $x+2=0$.
$$x=4$$ $$x=-2$$
Steps Using the Quadratic Formula
Combine $2x$ and $-4x$ to get $-2x$.
$$x^{2}-2x-8=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-2$ for $b$, and $-8$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
Now solve the equation $x=\frac{2±6}{2}$ when $±$ is plus. Add $2$ to $6$.
$$x=\frac{8}{2}$$
Divide $8$ by $2$.
$$x=4$$
Now solve the equation $x=\frac{2±6}{2}$ when $±$ is minus. Subtract $6$ from $2$.
$$x=-\frac{4}{2}$$
Divide $-4$ by $2$.
$$x=-2$$
The equation is now solved.
$$x=4$$ $$x=-2$$
Steps for Completing the Square
Combine $2x$ and $-4x$ to get $-2x$.
$$x^{2}-2x-8=0$$
Add $8$ to both sides. Anything plus zero gives itself.
$$x^{2}-2x=8$$
Divide $-2$, the coefficient of the $x$ term, by $2$ to get $-1$. Then add the square of $-1$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
$$x^{2}-2x+1=8+1$$
Add $8$ to $1$.
$$x^{2}-2x+1=9$$
Factor $x^{2}-2x+1$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.
$$\left(x-1\right)^{2}=9$$
Take the square root of both sides of the equation.
