Question

Factor the expression completely.

$x^2-2x-8$

Short answer

The solution$x^2-2x-8=\left(x-4\right)\left(x+2\right)$

Step-by-step solution

Step 1. Apply the concept of factoring trinomials.

To factor a trinomial of the form $x^2+bx+c$, we note that

$\left(x+r\right)\left(x+s\right)=x^2+\left(r+s\right)x+rs$

So, we need to choose numbers r and s so that$r+s=b$ and $rs=c$.

Step 2. Analyze the trinomial.

Given$x^2-2x-8$

This is of the form$x^2+bx+c$ where$b=-2$ and$c=-8$

The factorization of the trinomial will be$\left(x+r\right)\left(x+s\right)=x^2+\left(r+s\right)x+rs$

So, we need to choose numbers r and s so that$r+s=b$ and $rs=c$.

Step 3. Find the integers to factor the trinomial.

We need to choose integers r and s so that$r+s=-2$ and $rs=-8$.

By trial and error, we find that the two integers are $r=-4,s=2$

$r+s=-4+2=-2$

$rs=(-4)\cdot 2=-8$

Step 4. Factor the trinomial.

The factorization of the trinomial is

$\begin{array}{l}{x}^2-2x-8=\left(x+r\right)\left(x+s\right)\\ \Rightarrow x^2-2x-8=\left(x-4\right)\left(x+2\right)\end{array}$