Answer
$x=5$ or $x=12.3$
Work Step by Step
$V(x)=x(x+10)(30-2x)$,
$V(x)=x^2+10x(30-2x)$,
$V(x)=30x^2-2x^3+300x-20x^2$,
$V(x)=-2x^3+10x^2+300x=1500$,
$-2x^3+10x^2+300x-1500=0$, dividing both sides by $-2$.
$x^3-5x^2-150x+750=0$,
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, $a_{n}$.
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$f(x)=x^{3}-5x^{2}-150x+750$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2, \pm3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 25, \pm 30, \pm 50, \pm 75, \pm 125, \pm 150, \pm 250, \pm 375, \pm 750$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2, \pm3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 25, \pm 30, \pm 50, \pm 75, \pm 125, \pm 150, \pm 250, \pm 375, \pm 750$
b. Try for $x=5:$
$\begin{array}{lllll}
\underline{5}| & 1 & -5 & -150 & 750\\
& & 5 & 0 & -750\\
& -- & -- & -- & --\\
& 1 & 0 & -150 & |\underline{0}
\end{array}$
$5$ is a zero,
$f(x)=(x-5)(x^{2} -150)$
$f(x)=(x-5)(x^{2} -150)=0$,
$x=5$ or $x^2=150$,
$x= \pm 5\sqrt 6$.
since we are solving for depth (height), only the positive solution are relevant. Therefore, $x=5$ or $x=12.3$
