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Q. 27

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Found in: Page 241

### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

# Solve the inequality algebraically ${x}^{3}-2{x}^{2}-3x>0$

Required solution set is $\left(-1,0\right)\cup \left(3,\infty \right)$

See the step by step solution

## Step 1. Given information

we have a given inequality

${x}^{3}-2{x}^{2}-3x>0$

## Step 2.Finding the zeros

Zeros of inequality

$f\left(x\right)={x}^{3}-2{x}^{2}-3x>0are\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}f\left(x\right)=0\phantom{\rule{0ex}{0ex}}{x}^{3}-2{x}^{2}-3x=0\phantom{\rule{0ex}{0ex}}x\left({x}^{2}-2x-3\right)=0\phantom{\rule{0ex}{0ex}}x\left(x-3\right)\left(x+1\right)=0\phantom{\rule{0ex}{0ex}}x=0\phantom{\rule{0ex}{0ex}}x=3\phantom{\rule{0ex}{0ex}}x=-1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

## Step 3.Divide the real number line into 4 intervals

Now we use the zeros to separate the real number line into intervals.

$\left(-\infty ,-1\right)\left(-1,0\right)\left(0,3\right)\left(3,\infty \right)$

## Step 4.Selecting a test number in each interval

Now we select a test number in each interval found in Step 3 and evaluate at each number to determine if $f\left(x\right)={x}^{3}-2{x}^{2}-3x=0$ is positive or negative.

In the interval $\left(-\infty ,-1\right)\phantom{\rule{0ex}{0ex}}$we chose -2 where f is negative

In the interval $\left(-1,0\right)$we chose -0.5 , where f is positive.

In the interval (0,3) we chose 2.5 , where f is negative.

In the interval $\left(3,\infty \right)$we chose 4 , where f is positive.

We know that our required inequality is $f\left(x\right)>0$

Here the inequality is not strict $\left(\ge or\le \right)$ so we have to exclude the solutions of $f\left(x\right)=0$ in the solution set.

So we want the interval where f is positive.

So the required solution set is $\left(-1,0\right)\cup \left(3,\infty \right)$