已知 $2x^4 - kx^3 - 2kx^2 + 5x + 6$ 除以 $(2x - k)$ 的余数为 $7$。

1. Find the value of $x$ corresponding to the divisor $2x-k$.

For divisor $2x-k$,
$2x-k=0$
$x=\frac{k}{2}.$

2. Apply the Remainder Theorem and simplify to obtain the required equation.

Let
$P(x)=2x^4-kx^3-2kx^2+5x+6.$
Since the remainder is $7$,
$P\left(\frac{k}{2}\right)=7.$
So
$2\left(\frac{k}{2}\right)^4-k\left(\frac{k}{2}\right)^3-2k\left(\frac{k}{2}\right)^2+5\left(\frac{k}{2}\right)+6=7.$
Simplifying,
$\frac{k^4}{8}-\frac{k^4}{8}-\frac{k^3}{2}+\frac{5k}{2}+6=7.$
$-\frac{k^3}{2}+\frac{5k}{2}-1=0.$
Multiplying by $-2$,
$k^3-5k+2=0.$

1. Use the equation from part (i) and find one simple root.

From part (i),
$k^3-5k+2=0.$
Testing $k=2$,
$2^3-5(2)+2=8-10+2=0.$
So $k=2$ is a root, and $(k-2)$ is a factor.

2. Factorise the cubic using the known factor.

$k^3-5k+2=(k-2)(k^2+2k-1).$
Hence,
$(k-2)(k^2+2k-1)=0.$

3. Solve the remaining quadratic factor.

$k^2+2k-1=0$
Using the quadratic formula,
$k=\frac{-2\pm\sqrt{2^2-4(1)(-1)}}{2(1)}$
$k=\frac{-2\pm\sqrt8}{2}$
$k=-1\pm\sqrt2.$
Together with $k=2$, the values of $k$ are
$k=2,\;-1+\sqrt2,\;-1-\sqrt2.$

1. Rewrite the equation so that it matches the equation solved in part (ii).

Since $y=0$ is not a solution,
$2y^3-5y^2+1=0$
Divide by $y^3$:
$2-\frac5y+\frac1{y^3}=0.$
Let $k=\dfrac1y$. Then
$k^3-5k+2=0.$

2. Use the known values of $k$ from part (ii) to find the corresponding values of $y$.

From part (ii),
$k=-2,\;\frac{1+\sqrt5}{2},\;\frac{1-\sqrt5}{2}.$
Since $k=\dfrac1y$, $y=\dfrac1k$.

So
$y=-\frac12,$
$y=\frac{2}{1+\sqrt5}=\frac{\sqrt5-1}{2}=\frac{1}{\frac{1+\sqrt5}{2}},$
$y=\frac{2}{1-\sqrt5}=-\frac{1+\sqrt5}{2}=\frac{1-\sqrt5}{2}.$
Equivalently, the two reciprocal golden-ratio values are $\dfrac{1+\sqrt5}{2}$ and $\dfrac{1-\sqrt5}{2}$.