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

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

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.