Answer
$x=3$
Work Step by Step
$V_1(x)=(x+2)(2x+1)(x+5)$,
$V_1(x)=2x^2+x+4x+2(x+5)$,
$V_1(x)=2x^3+15x^2+27x+10$, is the volume of the cube with the missing part.
$V_2(x)=3x(x+5)$,
$V_2(x)=3x^2+15x$, is the volume of the missing part.
The volume of the solid shown in the figure is the volume of the cube with missing part minus the volume of of the missing part.
$V(x)=V_1(x)-V_2(x)$
$V(x)=2x^3+15x^2+27x+10-3x^2-15x$,
$V(x)=2x^3+12x^2+12x+10=208$,
$2x^3+12x^2+12x-198=0$, diving both sides by $2$
$x^3+6x^2+6x-99=0$,
To solve the equation, we use possible rational solution to list out all the possible solutions and use synthetic division to divide the quadrinomial and factor out.
See The Rational Zero Theorem:
If $\frac{p}{q}$ is a zero of the polynomial $f(x) $with integer coefficients, then $p$ is a factor of the constant term, $a_{0}$, and $q$ is a factor of the leading coefficient, $a_{n}$.
------------------------
$f(x)=x^3+6x^2+6x-99$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1,\pm3, \pm 9, \pm 11, \pm 33, \pm 99$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 3, \pm 9, \pm 11, \pm 33, \pm 99$
b. Try for $x=3:$
$\begin{array}{lllll}
\underline{3}| & 1 & 6 & 6 & -99\\
& & 3 & 27 & 99\\
& -- & -- & -- & --\\
& 1 & 9 & 33 & |\underline{0}
\end{array}$
$3$ is a zero,
$f(x)=(x-3)(x^{2} +9x+33)$
$f(x)=(x-3)(x^{2}+9x+33)=0$,
Solving for trinomial, since the determinant, $b^2-4ac$ is less than $0$ for the trinomial, only $x=3$ is the real solution to the equation.
