## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 1 - Section 1.8 - Inequalities - 1.8 Exercises - Page 89: 56

#### Answer

The solution is $[-3,0]\cup[0,3]$ or simply $[-3,3]$ The graph is:

#### Work Step by Step

$4x^{2}(x^{2}-9)\le0$ Factor the left side completely: $4x^{2}(x-3)(x+3)\le0$ Find the intervals. The factors are $x^{2}$, $x-3$ and $x+3$. Set them equal to $0$ and solve for $x$: $x^{2}=0$ $x=0$ $x-3=0$ $x=3$ $x+3=0$ $x=-3$ The factors are zero when $x=0,3,-3$. These three numbers divide the real line into the following intervals: $(-\infty,-3)$ $,$ $(-3,0)$ $,$ $(0,3)$ $,$ $(3,\infty)$ Elaborate a diagram, using test points to determine the sign of each factor in each interval: (refer to the attached image below) It can be seen from the diagram that the inequality is satisfied on the intervals $(-3,0)$ and $(0,3)$. Also, the inequality involves $\le$ so the endpoints also satisfy the inequality. The solution is $[-3,0]\cup[0,3]$ or simply $[-3,3]$

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