Ans: (a) Yes (b) Yes
- When $f(x)$ is divided by $x-2$, the remainder is $-33$. Then we have
$\begin{array}{rcl}
f(2) & = & -33 \\
4(2)^3 – 5(2)^2 – 18(2) + c & = & -33 \\
c – 24 & = & -33 \\
c & = & -9
\end{array}$Therefore, $f(x)=4x^3-5x^2-18x-9$. Consider $f(-1)$, we have
$\begin{array}{rcl}
f(-1) & = & 4(-1)^3 – 5(-1)^2 -18(-1) – 9 \\
& = & 0
\end{array}$Hence by the factor theorem, $x+1$ is a factor of $f(x)$.
- By the result of (a), we have
$\begin{array}{rcl}
f(x) & = & 0 \\
4x^3 – 5x^2 – 18x – 9 & = & 0 \\
(x+1)(4x^2 – 9x – 9) & = & 0 \\
(x+1) (x-3) (4x + 3) & = & 0
\end{array}$Therefore, the roots of the equation are $x=-1$, $x=3$ and $x= \dfrac{-3}{4}$, which are all rational numbers. Hence, I agree.
