College Algebra, Chapter 2, 2.3, Section 2.3, Problem 62

Find the solutions of the inequality $16x^2 + 24x^2 > - 9x - 1$ by drawing appropriate graphs. State each answer correct to two decimal places.


Based from the graph, the solutions are approximately
$(-1, -0.25) \bigcup (-0.25, \infty)$

By using algebra,

$\begin{equation} \begin{aligned} 16x^3 + 24x^2 &> - 9x - 1 \\ \\ 16x^3 + 24x^2 + 9x + 1 &> 0 && \text{Add } 9x \text{ and } 1 \end{aligned} \end{equation}$

By using synthethic division,



We have,

$\begin{equation} \begin{aligned} (x+1)\left( 16x^2 + 8x + 1\right) &> 0\\ \\ (x+1)\left( x + \frac{1}{4} \right)^2 &> 0 && \text{Perfect square} \end{aligned} \end{equation}$


The factors on the left hand side are $x+1$ and $x + \frac{1}{4}$. These factor are zero when $x$ is -1 and $-\frac{1}{4}$, respectively. These factor divide the number line into intervals.
$(-\infty, -1)\left( -1, -\frac{1}{4} \right]\left( -\frac{1}{4}, \infty \right)$
By testing the inequality,



Thus, the solution set is...
$\displaystyle \left(-1,-\frac{1}{4} \right)\bigcup \left( -\frac{1}{4},\infty \right)$