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

### Step 1: Substitute \( x = i \) and \( x = -i \)

To find the coefficients \( a \) and \( b \), we can evaluate \( P(i) \) and \( P(-i) \):

1. **Calculate \( P(i) \)**:
\[ P(i) = i^{19} + i^{17} + i^{13} + i^{11} + i^{7} + i^{4} + i^{3} \]

Using the fact that \( i^2 = -1 \), we can simplify the powers of \( i \):
- \( i^3 = -i \)
- \( i^4 = 1 \)
- \( i^5 = i \)
- \( i^6 = -1 \)
- \( i^7 = -i \)
- \( i^8 = 1 \)
- \( i^9 = i \)
- \( i^{10} = -1 \)
- \( i^{11} = -i \)
- \( i^{12} = 1 \)
- \( i^{13} = i \)
- \( i^{14} = -1 \)
- \( i^{15} = -i \)
- \( i^{16} = 1 \)
- \( i^{17} = i \)
- \( i^{18} = -1 \)
- \( i^{19} = -i \)

Now substituting these values into \( P(i) \):
\[ P(i) = (-i) + i + i + (-i) + (-i) + 1 + (-i) \]
Simplifying this:
\[ P(i) = (-i + i + i - i - i) + 1 = -i + 1 = 1 - i \]

2. **Calculate \( P(-i) \)**:
\[ P(-i) = (-i)^{19} + (-i)^{17} + (-i)^{13} + (-i)^{11} + (-i)^{7} + (-i)^{4} + (-i)^{3} \]

Using the same simplifications:
- \( (-i)^{3} = i \)
- \( (-i)^{4} = 1 \)
- \( (-i)^{5} = -i \)
- \( (-i)^{6} = 1 \)
- \( (-i)^{7} = i \)
- \( (-i)^{8} = 1 \)
- \( (-i)^{9} = -i \)
- \( (-i)^{10} = 1 \)
- \( (-i)^{11} = i \)
- \( (-i)^{12} = 1 \)
- \( (-i)^{13} = -i \)
- \( (-i)^{14} = 1 \)
- \( (-i)^{15} = i \)
- \( (-i)^{16} = 1 \)
- \( (-i)^{17} = -i \)
- \( (-i)^{18} = 1 \)
- \( (-i)^{19} = i \)

Now substituting these values into \( P(-i) \):
\[ P(-i) = i + (-i) + (-i) + i + i + 1 + i \]
Simplifying this:
\[ P(-i) = (i - i - i + i + i) + 1 = i + 1 \]

### Step 2: Set up the equations

Now we have the system of equations based on the remainder \( R(x) = ax + b \):
1. \( ai + b = 1 - i \)
2. \( -ai + b = 1 + i \)

### Step 3: Solve the equations

From the first equation:
\[ ai + b = 1 - i \quad \text{(1)} \]
From the second equation:
\[ -ai + b = 1 + i \quad \text{(2)} \]

Subtracting equation (1) from equation (2):
\[ -ai + b - (ai + b) = (1 + i) - (1 - i) \]
This simplifies to:
\[ -2ai = 2i \implies a = -1 \]

Substituting \( a = -1 \) back into equation (1):
\[ -i + b = 1 - i \implies b = 1 \]

### Final Result

Thus, the remainder when \( P(x) \) is divided by \( x^2 + 1 \) is:
\[ R(x) = -x + 1 \]

### Summary

The remainder is \( -x + 1 \).