What is the remainder when the polynomial x¹⁰⁰- 2x⁵¹ + 1 is divided by by (x²-1) ?

Given:

Polynomial P(x) = x¹⁰⁰ - 2x⁵¹ + 1

Divisor = x² - 1

Formula used:

The Remainder Theorem states that if a polynomial P(x) is divided by (x - c), the remainder is P(c).

For division by a quadratic, we can use the division algorithm: P(x) = Q(x)D(x) + R(x), where D(x) is the divisor, Q(x) is the quotient, and R(x) is the remainder.

The degree of R(x) must be less than the degree of D(x). Since D(x) = x² - 1 (degree 2), the remainder R(x) will be of the form (Ax + B).

Calculation:

Let P(x) = x¹⁰⁰ - 2x⁵¹ + 1

The divisor is x² - 1. We can factor this as (x - 1)(x + 1).

Let the remainder be R(x) = Ax + B, since the divisor is a quadratic (degree 2), the remainder must be linear (degree 1) or a constant.

According to the division algorithm:

P(x) = Q(x)(x² - 1) + (Ax + B)

P(x) = Q(x)(x - 1)(x + 1) + (Ax + B)

Now, substitute the roots of the divisor (x² - 1 = 0 ⇒ x = ±1) into the equation.

When x = 1:

P(1) = (1)¹⁰⁰ - 2(1)⁵¹ + 1

P(1) = 1 - 2(1) + 1

P(1) = 1 - 2 + 1

P(1) = 0

Also, from P(x) = Q(x)(x - 1)(x + 1) + (Ax + B), substituting x = 1:

P(1) = Q(1)(1 - 1)(1 + 1) + (A(1) + B)

P(1) = Q(1)(0)(2) + A + B

P(1) = A + B

So, we have our first equation:

A + B = 0 ......(Equation 1)

When x = -1:

P(-1) = (-1)¹⁰⁰ - 2(-1)⁵¹ + 1

Since (-1)^{even number} = 1 and (-1)^{odd number} = -1:

P(-1) = 1 - 2(-1) + 1

P(-1) = 1 + 2 + 1

P(-1) = 4

Also, from P(x) = Q(x)(x - 1)(x + 1) + (Ax + B), substituting x = -1:

P(-1) = Q(-1)(-1 - 1)(-1 + 1) + (A(-1) + B)

P(-1) = Q(-1)(-2)(0) - A + B

P(-1) = -A + B

So, we have our second equation:

-A + B = 4 ......(Equation 2)

Now, solve the system of linear equations (Equation 1 and Equation 2):

A + B = 0

-A + B = 4

Add Equation 1 and Equation 2:

(A + B) + (-A + B) = 0 + 4

2B = 4

B = 4 / 2 = 2

Substitute B = 2 into Equation 1:

A + 2 = 0

A = -2

The remainder R(x) is Ax + B.

R(x) = -2x + 2 = 2(1 - x)

∴ The correct answer is option 4.