## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 1 - Section 1.8 - Inequalities - 1.8 Exercises - Page 89: 54

#### Answer

The solution is $(-1,\infty)$ The graph is:

#### Work Step by Step

$(x+3)^{2}(x+1)\gt0$ Begin immediately by finding the intervals, because all nonzero terms are on one side of the inequality and the nonzero side is given in factored form. The factors are $x+1$ and $(x+3)^{2}$. Set them equal to $0$ and solve for $x$: $x+1=0$ $x=-1$ $(x+3)^{2}=0$ $x+3=0$ $x=-3$ The factors are zero when $x=-1,-3$. These two numbers divide the real line into the following intervals: $(-\infty,-3)$ $,$ $(-3,-1)$ $,$ $(-1,\infty)$ Elaborate a diagram, using test points to determine the sign of each factor in each interval: (refer to the the attached image below) It can be seen from the diagram that the inequality is satisfied only on the interval $(-1,\infty)$. Also, the inequality involves $\gt$ so the endpoints don't satisfy the inequality. The solution is $(-1,\infty)$

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