Question

Solve each inequality algebraically.

$\frac{x(x^2+1)(x-2)}{(x-1)(x+1)}\ge 0$

Short answer

Solution to the inequality $\frac{x(x^2+1)(x-2)}{(x-1)(x+1)}\ge 0$is $(-\infty ,-1)\cup [0,1)\cup [2,\infty )$.

Step-by-step solution

Step 1. Given information  

We have been given an inequality $\frac{x(x^2+1)(x-2)}{(x-1)(x+1)}\ge 0$.

We have to solve this inequality algebraically.

Step 2. Determine the real numbers at which the expression f equals zero and at which the expression f is undefined.  

Assume $f\left(x\right)=\frac{x(x^2+1)(x-2)}{(x-1)(x+1)}$.

$$\begin{gathered} x=0 \\ {x}^2+1=0 \\ Thereisnosuchrealx. \\ x-2=0 \\ x=2 \\ x-1=0 \\ x=1 \\ x+1=0 \\ x=-1 \end{gathered}$$

Step 3. Form the intervals  

Using the values of x found in previous step, we can divide the real numbers in the intervals:

$(-\infty ,-1)\cup (-1,0)\cup (0,1)\cup (1,2)\cup (2,\infty )$

Step 4. Select a number in each interval and evaluate f at the number  

Create the following table:

Interval	$(-\infty ,-1)$	$(-1,0)$	$(0,1)$	$(1,2)$	$(2,\infty )$
Number chosen	$-2$	$-0.5$	$0.5$	$1.5$	4
Value of f	$f(-2)=\frac{40}{3}$	$f(-0.5)=-2.08$	$f(0.5)=1.25$	$f(1.5)=-1.95$	$f\left(4\right)=\frac{136}{15}$
Conclusion	positive	negative	positive	negative	positive

Step 5. Identify the interval 

Since we want to know where f is positive or zero, we conclude that $f\left(x\right)\ge 0$in the interval $(-\infty ,-1)\cup [0,1)\cup [2,\infty )$.