Chapter 5 Polynomials and Polynomial Functions - 5.4 Factor and Solve Polynomial Equations - 5.4 Exercises - Problem Solving - Page 359: 64b

$x=4$

Calculate $y^3+y^2$ for $y=1,2,...,10$ Solve $x^3+2x^2=96\\ \Leftrightarrow (\frac{x}{2})^3+(\frac{x}{2})^2=\frac{96}{8}=12$ Search for values of $12$ in the table. It matches $y=2$. Hence, $\frac{x}{2}=2 \rightarrow x=4$