- Note that the remainder is $0$. Then by factor theorem, we have
$\begin{array}{rcl}
f(2) & = & 0 \\
3(2)^3-7(2)^2 +k(2) – 8 & = & 0 \\
2k & = & 12 \\
k & = & 6
\end{array}$By long division, we have
$\require{enclose}\begin{array}{rl}
& \ \ 3x^2 -x \\
x – 2 & \enclose{longdiv}{3x^3 -7x^2+6x-8} \\
& \ \ \underline{3x^3-6x^2\phantom{00000000}} \\
& \ \ \ \ \ \ \ \ -x^2 + 6x \\
& \ \ \ \ \ \ \ \ \ \underline{-\ x^2 +2x\phantom{0000}} \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +4x – 8 \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{+\ 4x-8\ }
\end{array}$Hence, $f(x) \equiv (x-2)(3x^2-x+4)$.
Therefore, $a=3$, $b=-1$ and $c=4$.
- Consider the discriminant of $3x^2-x+4=0$, we have
$\begin{array}{rcl}
\Delta & = & (-1)^2 – 4(3)(4) \\
& = & -47 \\
& < & 0 \end{array}$Therefore, there is no real roots of the equation $3x^2-x+4=0$.
Hence, there is only one real roots of $f(x)=0$. Therefore, I don’t agree.
2013-I-12
Ans: (a) $a=3$, $b=-1$ and $c=4$ (b) No
