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Expert-verified Found in: Page 242 ### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465 # ${x}^{3}+2{x}^{2}-3x>0$

The required interval is $\left(-3,0\right)\cup \left(1,\infty \right)$

See the step by step solution

## Step 1. Given Information

${x}^{3}+2{x}^{2}-3x>0\phantom{\rule{0ex}{0ex}}$

This is given to us . we need to solve for x .

## Step 2. Factorization

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

The zeros of the equation are 0, 1 & -3

## Step 3.  Dividing the number line into intervals using the above zeros.

The intervals are :

$\left(-\infty ,-3\right)\phantom{\rule{0ex}{0ex}}\left(-3,0\right)\phantom{\rule{0ex}{0ex}}\left(0,1\right)\phantom{\rule{0ex}{0ex}}\left(1,\infty \right)$

## Step 4. Evaluating the value of function

$1.\left(-\infty ,-3\right)\phantom{\rule{0ex}{0ex}}letnumberbe-4\phantom{\rule{0ex}{0ex}}valueoffunctionat-4=-20\phantom{\rule{0ex}{0ex}}conditionnotsatisfied.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}2.\left(-3,0\right)\phantom{\rule{0ex}{0ex}}Letnumberbe-1\phantom{\rule{0ex}{0ex}}valueoffunctionat-1=4\phantom{\rule{0ex}{0ex}}Theconditionissatisfied.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}3.\left(0,1\right)\phantom{\rule{0ex}{0ex}}letnumberbe0.1\phantom{\rule{0ex}{0ex}}valueoffunctionat0.1=-0.099\phantom{\rule{0ex}{0ex}}conditionnotsatisfied.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}4.\left(0,\infty \right)\phantom{\rule{0ex}{0ex}}Letnumberbe2\phantom{\rule{0ex}{0ex}}valueoffunctionat2=10\phantom{\rule{0ex}{0ex}}Theconditionissatisfied.$

## Step 5. conclusion

The required interval is $\left(-3,0\right)\cup \left(1,\infty \right)$ ### Want to see more solutions like these? 