Ans: (a) $0$ (b) $(x+3)(2x-1)(3x-2)$
- It is given that $f(1)=4$. By the remainder theorem, the remainder of $f(x)\div(x-1)$ is $4$. By division algorithm, we have
$\begin{array}{rcl}
f(x) & = & (x-1)(6x^2+17x-2)+4 \\
& = & 6x^3+11x^2-19x+6 \\
f(-3) & = & 6(-3)^3+11(-3)^2-19(-3)+6 \\
& = & 0
\end{array}$ - Since $f(-3)=0$, then by the factor theorem, $(x+3)$ is a factor of $f(x)$.
$\require{enclose}\begin{array}{rl}
& \ \ 6x^2 – 7x + 2 \\
x +3 & \enclose{longdiv}{6x^3 +11x^2-19x+6} \\
& \ \ \underline{6x^3+18x^2\phantom{000000000}} \\
& \ \ \ \ \ \ \ \ -7x^2 \ \ – 19x \\
& \ \ \ \ \ \ \ \ \ \underline{-\ 7x^2\ \ -21x\phantom{000}\ } \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +2x + 6 \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{+\ 2x +6\ \ }
\end{array}$Hence, we have
$\begin{array}{rcl}
f(x) & = & 6x^3+11x^2-19x+6 \\
& = & (x+3)(6x^2-7x+2) \\
& = & (x+3)(2x-1)(3x-2)
\end{array}$
