# Chapter 8 - Section 8.5 - Polynomial and Rational Inequalities - Exercise Set - Page 649: 16

$$(-5,-\frac{1}{3})$$ Let $3x^2 + 16x \lt -5$ be a function $f$. Express the inequality in the form $f(x) \lt 0$. Begin by the rewriting the inequality so that $0$ is on the right side. $3x^2 + 16x \lt -5$ $3x^2 + 16x +5\lt -5+5$ $3x^2 + 16x +5\lt 0$ Find the $x$-intercepts by solving $3x^2 + 16x +5=0$ Factor: $3x^2 + 16x +5 = (3x+1)(x+5)$ $3x+1=0$ or $x+5=0$ $x = -\frac{1}{3}$ or $x = -5$ These $x$-intercepts serve as boundary points that separate the number line into intervals. The boundary points of this equation therefore divide the number line into three intervals: $(-∞, -5)(-5,-\frac{1}{3})(-\frac{1}{3}, +∞)$ Choose one test value within each interval and evaluate $f$ at that number. $(-∞, -5)$ Test Value = $-6$ $f=(3x+1)(x+5)$ $f=(3(-6)+1)(-6+5)$ $f=17$ Conclusion: $f (x) > 0$ for all $x$ in $(-∞, -5)$. $(-5, -\frac{1}{3})$ Test Value = $-1$ $f=(3x+1)(x+5)$ $f=(3(-1)+1)(-1+5)$ $f=-8$ Conclusion: $f (x) < 0$ for all $x$ in $(-5, -\frac{1}{3})$. $(-\frac{1}{3},+∞)$ Test Value = $0$ $f=(3x+1)(x+5)$ $f=(3(0)+1)(0+5)$ $f=5$ Conclusion: $f (x) > 0$ for all $x$ in $(-\frac{1}{3},+∞)$. Write the solution set, selecting the interval or intervals that satisfy the given inequality. We are interested in solving $f (x) \lt 0$, where $f (x) = 3x^2+16x+5$. Based on the solution above, $f(x)\lt0$ for all $x$ in $(-5,-\frac{1}{3})$. Thus, the solution set of the given inequality is: $$(-5,-\frac{1}{3})$$ 