21. solve the following. $x^{2}-2x - 8>0$. type a response

Explanation:

Step1: Factor the quadratic expression

$x^{2}-2x - 8=(x - 4)(x+2)$ So the inequality becomes $(x - 4)(x + 2)>0$.

Step2: Find the roots

Set $(x - 4)(x + 2)=0$. The roots are $x=4$ and $x=-2$.

Step3: Test intervals

We have three intervals to test: $x<-2$, $-2<x<4$, $x>4$. For $x<-2$, let $x=-3$. Then $(-3 - 4)(-3+2)=(-7)\times(-1) = 7>0$, so $x<-2$ is part of the solution. For $-2<x<4$, let $x = 0$. Then $(0 - 4)(0+2)=(-4)\times2=-8<0$, so $-2<x<4$ is not part of the solution. For $x>4$, let $x = 5$. Then $(5 - 4)(5+2)=1\times7 = 7>0$, so $x>4$ is part of the solution.

Answer:

$x<-2$ or $x>4$