Problem 97 - Intermediate Algebra

Question

Suppose that \(P(x)=x^{100}-x^{99}+x^{98}-x^{97}+\cdots+x^{2}-x+1\). Find the remainder when \(P(x)\) is divided by \(x-1\)

Step-by-Step Solution

Answer

The remainder when \(P(x)\) is divided by \(x-1\) is 1.

Step 1: Recognize the Remainder Theorem

The Remainder Theorem states that when a polynomial \(P(x)\) is divided by a linear factor \(x-a\), the remainder is \(P(a)\). In this problem, we are asked for the remainder when \(P(x)\) is divided by \(x-1\). Therefore, the remainder is \(P(1)\).

Step 2: Evaluate P(1)

To find \(P(1)\), substitute 1 for \(x\) in the polynomial: \(P(1) = 1^{100} - 1^{99} + 1^{98} - 1^{97} + \cdots + 1^2 - 1 + 1\). Since each power of 1 is simply 1, the expression simplifies to \(1 - 1 + 1 - 1 + \cdots + 1 - 1 + 1\).

Step 3: Count the Terms

Notice that \(P(x)\) is an alternating series starting with a positive term and consists of 100 terms (from \(x^{100}\) to \(x^0\)). This means there are an equal number of positive and negative terms, specifically 50 of each.

Step 4: Simplify the Alternating Series

Since the series alternates with equal positive and negative terms, all negative positive terms cancel each out except for the last positive "+1" from \(x^0\). Thus, \(P(1) = 1\).

Key Concepts

Polynomial Division

Polynomial division is an important concept in algebra that involves dividing one polynomial by another. In this process, we determine how many times a divisor goes into the dividend, often resulting in a quotient and a remainder. The divisor is usually a linear polynomial in the form of \(x-a\). When we divide \(P(x)\) by \(x-1\) in the problem provided, the primary goal is to find the remainder, not the complete quotient.

• To perform polynomial division, we can use synthetic or long division, but since we intend only to find the remainder when dividing by a linear term, the Remainder Theorem presents an efficient alternative.
• The Remainder Theorem specifies that the remainder of a polynomial \(P(x)\) divided by \(x-a\) is simply \(P(a)\).

This theorem makes it easier to solve problems efficiently by focusing on evaluating polynomials, bypassing more complex division procedures that might otherwise be necessary.

Evaluating Polynomials

Evaluating polynomials involves calculating a polynomial's value at a specific point. This concept is central to finding remainders in polynomial division through the Remainder Theorem.

• When evaluating a polynomial \(P(x)\) at \(x=a\), substitute \(a\) into every instance of \(x\) within the polynomial.
• For example, in the exercise, we evaluate \(P(1)\) by substituting 1 into the polynomial \(P(x) = x^{100} - x^{99} + \cdots + x^2 - x + 1\).

Upon substituting 1 for \(x\), each power of 1 remains 1, leading to a series of alternating terms: \(1 - 1 + 1 - 1 + \cdots + 1 - 1 + 1\).
This simplification is key to efficiently finding the remainder in polynomial problems, especially as demonstrated in algebraic contexts where rapid calculations are beneficial.

Alternating Series

An alternating series is a sequence of numbers where terms switch between positive and negative. Recognizing and simplifying alternating series can streamline calculations, especially when working with specific polynomial problems.

• In the given problem, the series starts with a positive term and continues to alternate through 100 terms.
• By categorizing each term as either positive or negative, we see that there are 50 positive terms and 50 negative terms.

For the polynomial \(P(x)\) evaluated at \(x=1\), all negative terms \(-1\) pair with positive terms \(+1\), canceling each other except for the final \(+1\).
This results in a total value of \(P(1) = 1\), which is the remainder when \(P(x)\) is divided by \(x-1\). Understanding how alternating terms don't result in complex calculations allows for efficient problem-solving and demonstrates the effectiveness of analyzing series within polynomial functions.