2023-I-13

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

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

$\{\begin{array}{ll}p=3& \dots \text{①}\\ 5q-12p=-16& \dots \text{②}\end{array}$

Sub. $\text{①}$ into $\text{②}$, 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 $x={\displaystyle \frac{-4}{3}}$ or $x^2+5x-12=0$.

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

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

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

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