多項式，替換新變數

 

We are given the equation:

$$(x^2 - 3x)^2 - 14(x^2 - 3x) + 40 = 0$$

Step 1: Substituting a New Variable

Let:

$$y = x^2 - 3x$$

Then, our equation transforms into:

$$y^2 - 14y + 40 = 0$$

Step 2: Solve the Quadratic Equation

The quadratic equation:

$$y^2 - 14y + 40 = 0$$

can be solved using the quadratic formula:

$$y = \frac{-(-14) \pm \sqrt{(-14)^2 - 4(1)(40)}}{2(1)}$$ $$y = \frac{14 \pm \sqrt{196 - 160}}{2}$$ $$y = \frac{14 \pm \sqrt{36}}{2}$$ $$y = \frac{14 \pm 6}{2}$$

Step 3: Find Values of $y$

$$y = \frac{14 + 6}{2} = \frac{20}{2} = 10$$
$$y = \frac{14 - 6}{2} = \frac{8}{2} = 4$$

Thus, we have two equations to solve:

$$x^2 - 3x = 10$$
$$x^2 - 3x = 4$$

---

Step 4: Solve for $x$

Equation 1: $x^2-3x-10=0$

Factorizing:

$$(x - 5)(x + 2) = 0$$
$$x - 5 = 0 \quad \text{or} \quad x + 2 = 0$$
$$x = 5 \quad \text{or} \quad x = -2$$

Equation 2: $x^2-3x-4=0$

Factorizing:

$$(x - 4)(x + 1) = 0$$
$$x - 4 = 0 \quad \text{or} \quad x + 1 = 0$$
$$x = 4 \quad \text{or} \quad x = -1$$

---

Final Answer

$$\boxed{{\displaystyle x=5,-2,4,-1}}$$