ANSWER:
w = -5/2, w = 4 and w = -4
STEP-BY-STEP EXPLANATION:
We have the following equiation:
[tex]2w^3+5w^2-32w-80=0[/tex]We solve with the help of factoring by grouping
[tex]\begin{gathered} (2w^3+5w^2)+(-32w-80)=0 \\ w^2\cdot(2w+5)-16\cdot(2w+5)=0 \\ (2w+5)\cdot(w^2-16)=0 \\ 2w+5=0\rightarrow w=-\frac{5}{2} \\ (w^2-16)=0\rightarrow w^2=16\rightarrow w=\pm4\rightarrow w=4,w=-4 \end{gathered}[/tex]The solutions are w = -5/2, w = 4 and w = -4
