College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 1, Equations and Graphs - Section 1.7 - Solving Inequalities - 1.7 Exercises - Page 148: 38


$(-\infty, -3) \cup (-2, +\infty)$ Refer to the image below for the graph.

Work Step by Step

Factor the trinomial to obtain: $(x+2)(x+3)\gt 0$ Next, find the zeros of each factor. To find value/s of $x$ that will make each factor equal to zero, equate each factor to zero then solve each equation: $\begin{array}{ccc} &x+2=0 &\text{ or } &x+3=0 \\&x=-2 &\text{ or } &x=-3 \end{array}$ Next step is to find the intervals. The zeros $-3$ and $-2$ divide the number line into three intervals, namely: $(-\infty, -3), (-3, -2), \text{ and } (-2, +\infty)$. $\bf\text{Make a table of signs}.$ (refer to the attached image below) $\bf\text{Solve}$ From the table of signs, it can be seen that $(x+2)(x+3)\gt 0$, which is equivalent to $x^2+5x+6\gt 0$, in the intervals $(-\infty, -3)$ and $(-2, +\infty)$. The inequality involves strictly $\gt$ therefore the endpoints $-3$ and $-2$ are not part of the solution set. Thus, the solution set is: $(-\infty, -3) \cup (-2, +\infty)$. To graph this, plot holes (or hollow dots) at $-3$ and $-2$ then shade the region to the left of $-3$ and to the right of $-2$. (refer to the attached image in the answer part above)
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