2023-I-13

$\begin{array}{rcl}
h(x) & = & g(x)(px+q) +(px+q) \\
h(x) & = & (px+q)(g(x) +1) \\
h(x) & = & (px+q)(x^3+5x^2-12x-1+1) \\
h(x) & = & (px+q)(x^3+5x^2-12x) \\
h(x) & = & px^4+(5p+q)x^3+(5q-12p)x^2-12qx
\end{array}$

By comparing the coefficients of $x^4$ and $x^2$, we have

$\left\{ \begin{array}{ll}
p=3 & \ldots\unicode{x2460} \\
5q-12p = -16 & \ldots \unicode{x2461}
\end{array}\right.$

Sub. $\unicode{x2460}$ into $\unicode{x2461}$, we have

$\begin{array}{rcl}
5q-12(3) & = & -16 \\
5q & = & 20 \\
q & = & 4
\end{array}$

Therefore, the quotient is $3x=4$.

$\begin{array}{rcl}
(3x+4)(x^3+5x^2-12x) & = & 0 \\
x(3x+4)(x^2+5x-12) & = & 0
\end{array}$

$\therefore x=0$ or $x=\dfrac{-4}{3}$ or $x^2+5x-12=0$.

For $x^2+5x-12=0$, we have

$\begin{array}{rcl}
x & = & \dfrac{-5\pm\sqrt{5^2-4(1)(-12)}}{2(1)} \\
x & = & \dfrac{-5 \pm \sqrt{73}}{2} \text{, which are not rational numbers.}
\end{array}$

Note that only $x=0$ and $x=\dfrac{-4}{3}$ are rational numbers.

Therefore, the equation $h(x)=0$ has $2$ rational roots.