The remainder obtained when the polynomial `x+x^(3)+x^(9)+x^(27)+x^(81)+x^(243)` is divided by `x^(2)-1` is

Options

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

Explanation

To find the remainder when the polynomial \( P(x) = x + x^3 + x^9 + x^{27} + x^{81} + x^{243} \) is divided by \( x^2 - 1 \), we can use the Remainder Theorem. The remainder when a polynomial is divided by a quadratic polynomial \( x^2 - 1 \) will be a linear polynomial of the form \( R(x) = ax + b \).

### Step 1: Identify the roots of the divisor
The roots of \( x^2 - 1 \) are \( x = 1 \) and \( x = -1 \).

### Step 2: Evaluate the polynomial at the roots
We will evaluate \( P(x) \) at both roots.

1. **Evaluate at \( x = 1 \)**:
\[ P(1) = 1 + 1^3 + 1^9 + 1^{27} + 1^{81} + 1^{243} = 1 + 1 + 1 + 1 + 1 + 1 = 6 \]

2. **Evaluate at \( x = -1 \)**:
\[ P(-1) = -1 + (-1)^3 + (-1)^9 + (-1)^{27} + (-1)^{81} + (-1)^{243} \]
\[ = -1 - 1 - 1 - 1 - 1 - 1 = -6 \]

### Step 3: Set up the system of equations
Since the remainder is of the form \( R(x) = ax + b \), we can set up the following equations based on our evaluations:

1. \( R(1) = a(1) + b = 6 \) → \( a + b = 6 \) (Equation 1)
2. \( R(-1) = a(-1) + b = -6 \) → \( -a + b = -6 \) (Equation 2)

### Step 4: Solve the system of equations
We can solve these two equations simultaneously.

From Equation 1:
\[ b = 6 - a \]

Substituting \( b \) into Equation 2:
\[ -a + (6 - a) = -6 \]
\[ -2a + 6 = -6 \]
\[ -2a = -12 \]
\[ a = 6 \]

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

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

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

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