多項式，已知餘數，求商

 

We are given the function:

$$f(x) = (4x^2 - 2x + 1)(mx + n) + 4x - 9$$

with the conditions:

$$f\left(-\frac{1}{2}\right) = -5, \quad f(1) = -8.$$

We need to determine $m$ and $n$, and then compute $f(2)$.

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Step 1: Compute $f(-\frac{1}{2})$

Substituting $x = -\frac{1}{2}$:

$$4\left(-\frac{1}{2}\right)^2 - 2\left(-\frac{1}{2}\right) + 1 = 4\left(\frac{1}{4}\right) + 1 + 1 = 1 + 1 + 1 = 3.$$
$$f\left(-\frac{1}{2}\right) = (3)(m(-\frac{1}{2}) + n) + 4(-\frac{1}{2}) - 9$$
$$= 3\left(-\frac{m}{2} + n\right) - 2 - 9$$
$$= 3n - \frac{3m}{2} - 11$$

Setting this equal to $-5$:

$$3n - \frac{3m}{2} - 11 = -5$$
$$3n - \frac{3m}{2} = 6$$
$$3n = \frac{3m}{2} + 6$$
$$n = \frac{m}{2} + 2$$

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Step 2: Compute $f(1)$

Substituting $x = 1$:

$$4(1)^2 - 2(1) + 1 = 4 - 2 + 1 = 3.$$
$$f(1) = 3(m(1) + n) + 4(1) - 9$$
$$= 3(m + n) + 4 - 9$$
$$= 3m + 3n - 5$$

Setting this equal to $-8$:

$$3m + 3n - 5 = -8$$
$$3m + 3n = -3$$
$$m + n = -1$$

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Step 3: Solve for $m$ and $n$

We have the system:

1. $n = \frac{m}{2} + 2$
   
2. $m + n = -1$
   

Substituting equation (1) into equation (2):

$$m + \left(\frac{m}{2} + 2\right) = -1$$
$$m + \frac{m}{2} + 2 = -1$$

Multiply everything by 2 to clear the fraction:

$$2m + m + 4 = -2$$
$$3m = -6$$
$$m = -2$$

Substituting $m = -2$ into $n=\frac{m}{2}+\mathrm{2:}$

$$n = \frac{-2}{2} + 2 = -1 + 2 = 1$$

Thus, we have:

$$m = -2, \quad n = 1.$$

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Step 4: Compute $f(2)$

Substituting $x = 2$:

$$4(2)^2 - 2(2) + 1 = 16 - 4 + 1 = 13.$$
$$f(2) = (13)(m(2) + n) + 4(2) - 9$$
$$= 13(-2(2) + 1) + 8 - 9$$
$$= 13(-4 + 1) + 8 - 9$$
$$= 13(-3) + 8 - 9$$
$$= -39 + 8 - 9$$
$$= -40$$

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Final Answer

$$\boxed{{\displaystyle -40}}$$