Introduction to Polynomial Division and Remainder Theorem

In mathematics, particularly in algebra, the Remainder Theorem is a useful tool to find the remainder of a polynomial division. This article outlines a detailed method to determine the remainder when dividing a specific polynomial by another polynomial using both classical and modern techniques such as modular arithmetic. Specifically, we will explore how to find the remainder of P(x) x^{100} - x^{99} - x - 1 when divided by x^2 - 3x 2.

Step-by-Step Solution Using Standard Methods

To solve this problem, let's start by using the classical method which involves evaluating the polynomial at the roots of the divisor.

Evaluating at the Roots

The polynomial x^2 - 3x 2 can be factored into (x - 1)(x - 2). Therefore, we need to evaluate P(x) at x 1 and x 2.

Evaluation at x 1

Substitute x 1 into P(x):

[ P(1) 1^{100} - 1^{99} - 1 1 - 1 - 1 0 ]

Evaluation at x 2

Substitute x 2 into P(x):

[ P(2) 2^{100} - 2^{99} - 2 2^{99} - 2 ]

Now, let's express the remainder as R(x) ax b. Using the evaluations:

[ R(1) a(1) b 0 Rightarrow a b 0 Rightarrow b -a ] [ R(2) a(2) b 2^{99} - 2 Rightarrow 2a b 2^{99} - 2 ]

Substitute b -a into the second equation:

[ 2a - a 2^{99} - 2 Rightarrow a 2^{99} - 2 ]

Then, substituting a back to find b:

[ b -a -2^{99} 2 ]

Thus, the remainder R(x) is:

[ R(x) (2^{99} - 2)x - 2^{99} 2 ]

However, a cleaner and more concise representation is:

[ R(x) 2^{99} - 1x - 2^{99} ]

Therefore, the remainder when P(x) is divided by x^2 - 3x 2 is:

[ boxed{2^{99} - 1x - 2^{99}} ]

Alternative Approach Using Modular Arithmetic

A more elegant method involves the use of modular arithmetic. We start by noting that x^2 - 3x 2 is equivalent to 0, so x^2 equiv 3x - 2

Recursive Modulation

By repeatedly substituting, we can generate a pattern:

[ x^2 equiv 3x - 2 ] [ x^3 equiv 3x^2 - 3x 2 3(3x - 2) - 3x 2 7x - 6 ] [ x^4 equiv 3x^3 - 3x^2 2x 3(7x - 6) - 3(3x - 2) 2x 15x - 14 ]

From this, we conjecture that:

[ x^n equiv 2^n - 1x - 2^n - 2 ]

To verify this, assume it holds for n l. Then for n l 1:

[ x^{l 1} equiv (2^l - 1)x - (2^l - 2)2 2^{l 1} - 1x - 2^{l 1} - 2 ]

Thus, the conjecture is correct. Applying this to the original polynomial:

[ x^{100} equiv 2^{100} - 1x - 2^{100} - 2 ] [ x^{99} equiv 2^{99} - 1x - 2^{99} - 2 ]

Substituting back:

[ P(x) x^{100} - x^{99} - x - 1 equiv (2^{100} - 1x - 2^{100} - 2) - (2^{99} - 1x - 2^{99} - 2) - x - 1 ] [ equiv 2^{100} - 2^{99}x - 2^{100} - 2 - 2^{99} 2 - x - 1 2^{99} - 1x - 2^{99} - 1 ]

So, the remainder is:

[ boxed{2^{99} - 1x - 1} ]