If the polynomial `x^(19) + x^(17) +x^(13) +x^(11) +x^(7) +x^(4) +x^(3)` is divided by `(x^(2)+1)`, then the remainder is

Options

A. 1
B. `x^(2)+4`
C. `-x`
D. x

Answer

To find the remainder when the polynomial \( f(x) = x^{19} + x^{17} + x^{13} + x^{11} + x^{7} + x^{4} + x^{3} \) is divided by \( g(x) = x^2 + 1 \), we can use the Remainder Theorem. According to the theorem, the remainder of a polynomial \( f(x) \) when divided by a polynomial \( g(x) \) of degree \( n \) is a polynomial of degree less than \( n \). Since \( g(x) \) is of degree 2, the remainder will be of the form \( ax + b \).

### Step-by-step Solution:

1. **Identify the polynomial and divisor**:
- Let \( f(x) = x^{19} + x^{17} + x^{13} + x^{11} + x^{7} + x^{4} + x^{3} \)
- Let \( g(x) = x^2 + 1 \)

2. **Use the Remainder Theorem**:
- We need to evaluate \( f(x) \) at the roots of \( g(x) \). The roots of \( g(x) = 0 \) are \( x = i \) and \( x = -i \) (where \( i \) is the imaginary unit).

3. **Calculate \( f(i) \)**:
- Substitute \( x = i \) into \( f(x) \):
\[ f(i) = i^{19} + i^{17} + i^{13} + i^{11} + i^{7} + i^{4} + i^{3} \]
- Calculate the powers of \( i \):
- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = -i \)
- \( i^4 = 1 \)
- \( i^5 = i \)
- \( i^6 = -1 \)
- \( i^7 = -i \)
- \( i^8 = 1 \)
- The pattern repeats every 4 terms.

- Now calculate each term:
\[ i^{19} = i^{(4 \cdot 4 + 3)} = i^3 = -i \]
\[ i^{17} = i^{(4 \cdot 4 + 1)} = i^1 = i \]
\[ i^{13} = i^{(4 \cdot 3 + 1)} = i^1 = i \]
\[ i^{11} = i^{(4 \cdot 2 + 3)} = i^3 = -i \]
\[ i^{7} = i^{(4 \cdot 1 + 3)} = i^3 = -i \]
\[ i^{4} = 1 \]
\[ i^{3} = -i \]

- Now sum these values:
\[ f(i) = (-i) + i + i + (-i) + (-i) + 1 + (-i) \]
- Combine like terms:
\[ = 1 + (-i - i - i - i + i) = 1 - 3i \]

4. **Calculate \( f(-i) \)**:
- Substitute \( x = -i \) into \( f(x) \):
\[ f(-i) = (-i)^{19} + (-i)^{17} + (-i)^{13} + (-i)^{11} + (-i)^{7} + (-i)^{4} + (-i)^{3} \]
- Calculate the powers of \( -i \):
\[ (-i)^{19} = -i, \quad (-i)^{17} = -i, \quad (-i)^{13} = -i, \quad (-i)^{11} = -i, \quad (-i)^{7} = -i, \quad (-i)^{4} = 1, \quad (-i)^{3} = i \]
- Now sum these values:
\[ f(-i) = (-i) + (-i) + (-i) + (-i) + (-i) + 1 + i \]
- Combine like terms:
\[ = 1 - 4i \]

5. **Form the remainder polynomial**:
- The remainder polynomial can be expressed as \( R(x) = ax + b \).
- We have two equations:
\[ R(i) = ai + b = 1 - 3i \quad \text{(1)} \]
\[ R(-i) = -ai + b = 1 - 4i \quad \text{(2)} \]

6. **Solve the system of equations**:
- From (1): \( ai + b = 1 - 3i \)
- From (2): \( -ai + b = 1 - 4i \)

- Subtract (1) from (2):
\[ (-ai + b) - (ai + b) = (1 - 4i) - (1 - 3i) \]
\[ -2ai = -i \implies a = \frac{1}{2} \]

- Substitute \( a \) back into (1):
\[ \frac{1}{2}i + b = 1 - 3i \]
\[ b = 1 - 3i - \frac{1}{2}i = 1 - \frac{7}{2}i \]

7. **Final remainder**:
- Thus, the remainder is:
\[ R(x) = \frac{1}{2}x + \left(1 - \frac{7}{2}i\right) \]

### Final Answer:
The remainder when \( f(x) \) is divided by \( g(x) \) is \( \frac{1}{2}x + 1 - \frac{7}{2}i \).