Question

find the interval that answers the inequality: 24. $\frac{(x - 1)^2(x + 2)}{(x + 1)}>0$

Explanation:

Step1: Find the critical points

Set the numerator and denominator equal to zero.
For $(x - 1)^2(x + 2)=0$, we have $x = 1$ (with multiplicity 2) and $x=-2$. For $x + 1=0$, we have $x=-1$.

Step2: Analyze the sign of the function in intervals

The intervals are $(-\infty,-2)$, $(-2,-1)$, $(-1,1)$ and $(1,\infty)$.
For $x\in(-\infty,-2)$, let $x=-3$. Then $\frac{(-3 - 1)^2(-3 + 2)}{(-3+1)}=\frac{16\times(-1)}{-2}=8>0$.
For $x\in(-2,-1)$, let $x =-\frac{3}{2}$. Then $\frac{(-\frac{3}{2}-1)^2(-\frac{3}{2}+2)}{(-\frac{3}{2}+1)}=\frac{(\frac{25}{4})\times(\frac{1}{2})}{(-\frac{1}{2})}=-\frac{25}{4}<0$.
For $x\in(-1,1)$, let $x = 0$. Then $\frac{(0 - 1)^2(0 + 2)}{(0 + 1)}=2>0$.
For $x\in(1,\infty)$, let $x = 2$. Then $\frac{(2 - 1)^2(2 + 2)}{(2 + 1)}=\frac{4}{3}>0$.
Since $(x - 1)^2\geq0$ for all real $x$ and $(x - 1)^2 = 0$ when $x = 1$, and we want $\frac{(x - 1)^2(x + 2)}{(x + 1)}>0$.

Answer:

$(-\infty,-2)\cup(-1,1)\cup(1,\infty)$