多項式，求未定係數

 

Given Equation

$$2x^3 - 4x^2 - 5x + 1 = a(x - 1)(x + 2)(x + 3) + b(x - 1)(x - 2)(x - 3) + c(x + 2)(x - 3)(x + 1)$$

We will substitute three specific values of $x$ to solve for $a, b, c$.

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Step 1: Substituting $x=1$

Since $(x-1)=\mathrm{0,\; the\; terms\; involving }$$a$ and $\mathrm{b vanish,\; leaving:}$

$$f(1) = c(1 + 2)(1 - 3)(1 + 1)$$

Compute $f(1):$

$$2(1)^3 - 4(1)^2 - 5(1) + 1 = 2 - 4 - 5 + 1 = -6$$

Compute the right-hand side:

$$c(3)(-2)(2) = -12c$$

Thus, we solve:

$$-6 = -12c$$
$$c = \frac{1}{2}$$

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Step 2: Substituting $x=-2$

Since $(x+2)=\mathrm{0,\; the\; terms\; involving }$$a$ and $c$ vanish, leaving:

$$f(-2) = b(-2 - 1)(-2 - 2)(-2 - 3)$$

Compute $f(-2)$:

$$2(-2)^3 - 4(-2)^2 - 5(-2) + 1$$
$$= 2(-8) - 4(4) - 5(-2) + 1$$
$$= -16 - 16 + 10 + 1 = -21$$

Compute the right-hand side:

$$b(-3)(-4)(-5) = -60b$$

Thus, we solve:

$$-21 = -60b$$
$$b = \frac{7}{20}$$

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Step 3: Substituting $x=3$

Since $(x-3)=\mathrm{0,\; the\; terms\; involving }$$b$ and $c$ vanish, leaving:

$$f(3) = a(3 - 1)(3 + 2)(3 + 3)$$

Compute $f(3)$:

$$2(3)^3 - 4(3)^2 - 5(3) + 1$$
$$= 2(27) - 4(9) - 15 + 1$$
$$= 54 - 36 - 15 + 1 = 4$$

Compute the right-hand side:

$$a(2)(5)(6) = 60a$$

Thus, we solve:

$$4 = 60a$$
$$a = \frac{1}{15}$$

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Final Answer

$$\boxed{a = \frac{1}{15}, \quad b = \frac{7}{20}, \quad c = \frac{1}{2}}$$

This is the correct set of values for $a, b, c$. 😊