Cubic equation with one integral root

Question

The context of the following polynomial equation is plane geometry. A hexagon is inscribed in a circle, a pair of vertices are endpoints of a diameter of the circle, and $x$ is the diameter of the circle.

\begin{equation*} x^{4} - 174x^{2} - 308x = 0 , \end{equation*} or equivalently, \begin{equation*} x(x^{3} - 174x - 308) = 0 . \end{equation*}

How does one extract the integral root from the cubic equation $x^{3} - 174x - 308 = 0$?

Answer 1

The factors of $308$ are $1,2,4,7,11,14,22,28,44,77,154,308$.

Trial and error using the rational root test gives $(x-14)$ as a root and after which we have $$x^3-174x-308=(x-14)(x^2-160x+22)$$

Note that the discriminant $(-160)^2-4(1)(22)=25512$ which is not a square so there are no more integer roots.

Answer 2

If $$ x^3 - 174x -308 = 0 $$ then $$ x(x^2-174) = 308 . $$ If $x$ is an integer it's a factor of $308$. Then you can finish with trial and error. If nothing works there's no integer root.