-
- Since $x+3$ is a factor of $f(x)$, then we have
$\begin{array}{rcl}
f(-3) & = & 0 \\
4(-3)^3+k(-3)^2 -243 & = & 0 \\
9k – 351 & = & 0 \\
k & = & 39
\end{array}$ - By long division, we have
$\require{enclose}\begin{array}{rl}
& \ \ 4x^2+27x-81 \\
x + 3 & \enclose{longdiv}{4x^3 + 39x^2 +\ \ 0x-243} \\
& \ \ \underline{4x^3 + 12x^2 \phantom{0000000000}\ \ \ } \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ 27x^2 + \ \ 0x \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{27x^2 +81x\phantom{0000}\ \ \ \ } \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -81x – 243 \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{-\ 81x – 243} \\
\end{array}$Hence, we have
$\begin{array}{rcl}
f(x) & = & (x+3)(4x^2+27x-81) \\
& = & (x+3)(x+9)(4x-9)
\end{array}$
- Since $x+3$ is a factor of $f(x)$, then we have
-
- Let $C=k_1x^3 + k_2x^2$, where $k_1$ and $k_2$ are non-zero constants. When $x=5.5$, $C=7381$, we have
$\begin{array}{rcl}
k_1(5.5)^3 + k_2(5.5)^2 & = & 7381 \\
5.5k_1 + k_2 & = & 244 \ \ldots \unicode{x2460}
\end{array}$When $x=6$, $C=9072$, we have
$\begin{array}{rcl}
k_1(6)^3 + k_2(6)^2 & = & 9072 \\
6 k_1 + k_2 & = & 252 \ \ldots \unicode{x2461}
\end{array}$$\unicode{x2461} – \unicode{x2460}$, we have
$\begin{array}{rcl}
0.5k_1 & = & 8 \\
k_1 & = & 16
\end{array}$Sub. $k_1=16$ into $\unicode{x2461}$, we have
$\begin{array}{rcl}
6(16) + k_2 & = & 252 \\
k_2 & = & 156
\end{array}$Therefore, $C=16x^3 + 156x^2$.
- Sub. $C=972$ into $C=16x^3+156x^2$, we have
$\begin{array}{rcl}
16x^3 + 156x^2 & = & 972 \\
4x^3 + 39x^2 – 243 & = & 0 \\
(x+3)(x+9)(4x-9) & = & 0
\end{array}$Therefore, $x=-3$, $x=-9$ or $x=\dfrac{9}{4}$.
Since the length of a handicraft must be positive, the required length is $\dfrac{9}{4}\text{ cm}$.
- Let $C=k_1x^3 + k_2x^2$, where $k_1$ and $k_2$ are non-zero constants. When $x=5.5$, $C=7381$, we have
2007-I-14
Ans: (a) (i) $39$ (ii) $(x+3)(x+9)(4x-9)$ (b) (i) $C=16x^3+156x^2$ (ii)$\dfrac{9}{4}\text{ cm}$
