Answer
$$(-5,-\frac{1}{3})$$
Work Step by Step
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:
$$(-5,-\frac{1}{3})$$
