College Algebra 7th Edition

$P(x)=\dfrac{1}{8}(2x+3)^2(x-2)^2(x^2+3x+9)^2$ Zeros: $-\dfrac{3}{2}$ and $2$ Refer to the graph below.
Factor the polynomial completely to obtain: \begin{align*} P(x)&=\frac{1}{8}\left[(2x^4+3x^3)+(-16x-24)\right]^2\\ P(x)&=\frac{1}{8}\left[x^3(2x+3)+(-8)(2x+3)\right]^2\\ P(x)&=\frac{1}{8}\left[(2x+3)(x^3-8)\right]^2\\ P(x)&=\frac{1}{8}\left[(2x+3)(x^3-8)\right]^2\\ P(x)&=\frac{1}{8}\left[(2x+3)(x-2)(x^2+3x+9)\right]^2\\ P(x)&=\frac{1}{8}(2x+3)^2(x-2)^2(x^2+3x+9)^2\\ \end{align*} To find the zeros, use the Zero-Product Property by equating each factor to $0$, then solve each equation to obtain: \begin{align*} (2x+3)^2&=0 &\text{or}& &(x-2)^2=0& &\text{or}& &(x^2+3x+9)^2=0\\ 2x+3&=0 &\text{or}& &x-2=0& &\text{or}& &x^2+3x+9=0\\ 2x&=-3 &\text{or}& &x=2& &\text{or}& &\text{ (no real solution)}\\ x&=-\frac{3}{2} \end{align*} Both real roots have multiplicity $2$. Use a graphing utility to graph $P(x)$. Refer to the graph above.