2022-I-14

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

By comparing the coefficients, we have

$\{\begin{array}{ll}c=2& \dots \text{①}\\ d-2c=a& \dots \text{②}\\ 3c-2d+1=b& \dots \text{③}\\ 3d+13=-20& \dots \text{④}\end{array}$

From $\text{①}$ and $\text{④}$, we have $c=2$ and $d=-11$. Then from $\text{②}$, we have

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

From $\text{③}$, 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

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

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