2022-I-14

$\begin{array}{rcl}
p(x) & = & (x^2-2x+3)(cx+d) +x +13\\
p(x) & = & cx^3 -2cx^2+3cx +dx^2-2dx +3d +x +13 \\
p(x) & = & cx^3 +(d-2c)x^2 +(3c -2d+1)x +3d +13
\end{array}$

By comparing the coefficients, we have

$\left\{ \begin{array}{ll}
c =2 & \ldots \unicode{x2460} \\
d-2c = a & \ldots \unicode{x2461} \\
3c -2d+1=b & \ldots \unicode{x2462} \\
3d +13 =-20 & \ldots \unicode{x2463}
\end{array}\right.$

From $\unicode{x2460}$ and $\unicode{x2463}$, we have $c=2$ and $d=-11$. Then from $\unicode{x2461}$, we have

$\begin{array}{rcl}
a & = & -11 -2(2) \\
a & = & -15
\end{array}$

From $\unicode{x2462}$, we have

$\begin{array}{rcl}
b & = & 3(2) -2(-11) +1 \\
b & = & 29
\end{array}$

$\begin{array}{rcl}
p(5) & = & 2(5)^3-15(5)^2 +29(5) -20 \\
p(5) & = & 0
\end{array}$

Therefore by the factor theorem, $x-5$ is a factor or $p(x)$.

$\therefore x=5$ or $2x^2 -5x +4=0$.

Consider $2x^2 -5x +4=0$, we have

$\begin{array}{rcl}
x & = & \dfrac{-(-5) \pm\sqrt{(-5)^2 -4(2)(4)}}{2(2)} \\
x & = & \dfrac{5 \pm \sqrt{-7}}{4} \text{, which are complex numbers.}
\end{array}$

Since $5$ and $\dfrac{5\pm\sqrt{-7}}{4}$ are not irrational numbers, then the claim is disagreed.