The remainder obtained when the polynomial `x+x^(3)+x^(9)+x^(27)+x^(81)+x^(243)` is divided by `x^(2)-1` is (a)6x+1 (b)5x+1 (c)4x (d)6x

Options

A. `6x+1`
B. `5x+1`
C. `4x`
D. `6x`

Explanation

To find the remainder of the polynomial \( P(x) = x + x^3 + x^9 + x^{27} + x^{81} + x^{243} \) when divided by \( x^2 - 1 \), we can use the Remainder Theorem. The theorem states that the remainder of a polynomial \( P(x) \) when divided by \( x - c \) is \( P(c) \). Since \( x^2 - 1 = (x - 1)(x + 1) \), we will evaluate the polynomial at \( x = 1 \) and \( x = -1 \).

### Step 1: Evaluate \( P(1) \)
Substituting \( x = 1 \) into the polynomial:
\[ P(1) = 1 + 1^3 + 1^9 + 1^{27} + 1^{81} + 1^{243} \]
Calculating each term:
\[ P(1) = 1 + 1 + 1 + 1 + 1 + 1 = 6 \]

### Step 2: Evaluate \( P(-1) \)
Substituting \( x = -1 \) into the polynomial:
\[ P(-1) = -1 + (-1)^3 + (-1)^9 + (-1)^{27} + (-1)^{81} + (-1)^{243} \]
Calculating each term:
\[ P(-1) = -1 - 1 - 1 - 1 - 1 - 1 = -6 \]

### Step 3: Form the remainder polynomial
The remainder when dividing by \( x^2 - 1 \) can be expressed in the form \( R(x) = ax + b \). We have two equations from our evaluations:
1. \( R(1) = a(1) + b = 6 \)
2. \( R(-1) = a(-1) + b = -6 \)

This gives us the system of equations:
1. \( a + b = 6 \) (Equation 1)
2. \( -a + b = -6 \) (Equation 2)

### Step 4: Solve the system of equations
Subtract Equation 2 from Equation 1:
\[ (a + b) - (-a + b) = 6 - (-6) \]
This simplifies to:
\[ 2a = 12 \implies a = 6 \]

Substituting \( a = 6 \) back into Equation 1:
\[ 6 + b = 6 \implies b = 0 \]

### Step 5: Write the final remainder
Thus, the remainder \( R(x) \) is:
\[ R(x) = 6x + 0 = 6x \]

### Conclusion
The remainder obtained when the polynomial \( x + x^3 + x^9 + x^{27} + x^{81} + x^{243} \) is divided by \( x^2 - 1 \) is \( 6x \).

### Answer
The correct option is (d) \( 6x \).