2021-I-12

$p(x)=(x^2+x+1)(2x^2-37)+cx+c-1$

Since $p(x)$ is divisible by $x-5$, we have

$\begin{array}{rcl}p(5)& =& 0\\ (5^2+5+1)(2(5{)}^2-37)+c(5)+c-1& =& 0\\ 403+6c–1& =& 0\\ c& =& -67\end{array}$

$\begin{array}{cl}& p(x)\\ =& (x^2+x+1)(2x^2-37)-67x-67-1\\ =& (x^2+x+1)(2x^2-37)-67x-68\\ =& 2x^4+2x^3-35x^2-37x-37-67x-68\\ =& 2x^4+2x^3-35x^2-104x-105\end{array}$

Consider $p(-3)$, we have

$\begin{array}{cl}& p(-3)\\ =& 2(-3{)}^4+2(-3{)}^3-35(-3{)}^2-104(-3)-105\\ =& 0\end{array}$

Therefore by the factor theorem, $x+3$ is a factor of $p(x)$.

Consider the discriminant of $2x^2+6x+7=0$, we have

$\begin{array}{cl}& \mathrm{\Delta }\\ =& 6^2–4(2)(7)\\ =& -20\\ <& 0\end{array}$

Therefore, there is no real roots for the equation $2x^2+6x+7=0$.

Hence, not all the roots of $p(x)=0$ are real numbers. The claim is not correct.