使用长除法，用 $2x^{3}-7x^{2}+2x+3$ 除以 $2x-4$。 解方程 $2x^{3}-7x^{2}+2x+3=0$。 Kun Ye 声称方程 $2x^{3}-7x^{2}+2x+3=0$ 的解可以用来求解 $3y^{6}+2y^{4}-7y^{2}+2=0$。解释为什么他是正确的。

1. Set up the long division and find the first term of the quotient.

Divide the leading term $2x^3$ by $2x$ to get $x^2$.
Multiply $x^2$ by $(2x-4)$ to get $2x^3 - 4x^2$.
Subtract this from $2x^3 - 7x^2$ to get $-3x^2$, and bring down the next term $2x$.

2. Find the second term of the quotient.

Divide $-3x^2$ by $2x$ to get $-\frac{3}{2}x$.
Multiply $-\frac{3}{2}x$ by $(2x-4)$ to get $-3x^2 + 6x$.
Subtract this from $-3x^2 + 2x$ to get $-4x$, and bring down the $3$.

3. Find the final term of the quotient and the remainder.

Divide $-4x$ by $2x$ to get $-2$.
Multiply $-2$ by $(2x-4)$ to get $-4x + 8$.
Subtract this from $-4x + 3$ to get a remainder of $-5$.

1. Use the Factor Theorem to find a linear factor of the polynomial.

Let $f(x) = 2x^{3} - 7x^{2} + 2x + 3$.
$f(1) = 2(1)^{3} - 7(1)^{2} + 2(1) + 3 = 2 - 7 + 2 + 3 = 0$
Since $f(1) = 0$, $(x - 1)$ is a factor.

2. Find the quadratic factor by dividing the cubic polynomial by the linear factor.

$2x^{3} - 7x^{2} + 2x + 3 = (x - 1)(ax^{2} + bx + c)$
By inspection or long division:
$a = 2$ and $c = -3$.
Comparing coefficients of $x^{2}$:
$-a + b = -7 \implies -2 + b = -7 \implies b = -5$
So, the quadratic factor is $2x^{2} - 5x - 3$.

3. Factorise the quadratic factor completely.

$2x^{2} - 5x - 3 = (2x + 1)(x - 3)$

4. Solve the equation to find all values of $x$.

$(x - 1)(2x + 1)(x - 3) = 0$
$x - 1 = 0 \implies x = 1$
$2x + 1 = 0 \implies x = -\frac{1}{2}$
$x - 3 = 0 \implies x = 3$

1. Identify the relationship between the two equations by substituting $x = \frac{1}{y^2}$

Let $x = \frac{1}{y^2}$.
Substitute into $2x^3 - 7x^2 + 2x + 3 = 0$:
$2\left(\frac{1}{y^2}\right)^3 - 7\left(\frac{1}{y^2}\right)^2 + 2\left(\frac{1}{y^2}\right) + 3 = 0$
$\frac{2}{y^6} - \frac{7}{y^4} + \frac{2}{y^2} + 3 = 0$

2. Multiply by $y^6$ to obtain the second equation

Multiply the entire equation by $y^6$:
$2 - 7y^2 + 2y^4 + 3y^6 = 0$
Rearranging the terms gives:
$3y^6 + 2y^4 - 7y^2 + 2 = 0$
This matches the second equation, showing that if $x$ is a solution to the first equation, then $y$ where $y^2 = \frac{1}{x}$ is a solution to the second.