College Algebra, Chapter 4, 4.2, Section 4.2, Problem 36

Factor the polynomial $\displaystyle P(x) = \frac{1}{8} \left( 2x^4 + 3x^3 - 16x - 24 \right)^2$ and use the factored form to find the zeros. Then sketch the graph.
Since the function has an even degree of 8 and a positive leading coefficient. Its end behaviour is $y \rightarrow \infty \text{ as } x \rightarrow - \infty \text{ and } y \rightarrow \infty \text{ as } x \rightarrow \infty$. To solve for the $x$-intercepts (or zeros), we set $y = 0$

$
\begin{equation}
\begin{aligned}
0 &= \frac{1}{8} \left( 2x^4 + 3x^3 - 16x - 24 \right)^2\\
\\
0 &= \left( 2x^4 + 3x^3 - 16x - 24 \right)^2 && \text{Multiply by 8}\\
\\
0 &= \left[ 2x^4 + 3x^3 - (16x + 24) \right]^2 && \text{Group terms}\\
\\
0 &= \left[ x^3 (2x+3) - 8 (2x+3) \right]^2 && \text{Factor out } x^3 \text{ and } 8\\
\\
0 &= \left[ (x^3 - 8)(2x +3) \right]^2 && \text{Factor out } 2x + 3\\
\\
0 &= (x^3 -8)^2 (2x + 3)^2 && \text{Simplify}
\end{aligned}
\end{equation}
$

To find the real zeros of $P$, we set $x^3 - 8 = 0$ and $2x + 3 = 0$. So $x = 2$ and $\displaystyle \frac{-1}{2}$ are the zeros of $P$. The two zeros of $P$ has a multiplicity of 2, it means that the graph bounce off at these zeros.