Problem 54 - College Algebra with Corequisite Support

Question

For the following exercises, use a calculator with CAS to answer the questions. Consider \(\frac{x^{k}-1}{x-1}\) with \(k=1,2,3\). What do you expect the result to be if \(k=4\) ?

Step-by-Step Solution

Answer

For \(k=4\), the result is expected to be \(x^3 + x^2 + x + 1\).

Step 1: Analyze the Given Expression

The given expression is \(\frac{x^k - 1}{x - 1}\). We need to evaluate this for different values of \(k\): specifically \(k = 1, 2, 3\), and then predict for \(k = 4\).

Step 2: Calculate for k=1

Substitute \(k = 1\) into the expression: \(\frac{x^1 - 1}{x - 1} = \frac{x - 1}{x - 1}\). Simplifying, this gives \(1\), provided \(x eq 1\).

Step 3: Calculate for k=2

Substitute \(k = 2\) into the expression: \(\frac{x^2 - 1}{x - 1}\). The numerator \(x^2 - 1\) can be factored to \((x - 1)(x + 1)\). Thus, \(\frac{(x - 1)(x + 1)}{x - 1} = x + 1\), assuming \(x eq 1\).

Step 4: Calculate for k=3

Substitute \(k = 3\) into the expression: \(\frac{x^3 - 1}{x - 1}\). The numerator \(x^3 - 1\) can be factored to \((x - 1)(x^2 + x + 1)\). Thus, \(\frac{(x - 1)(x^2 + x + 1)}{x - 1} = x^2 + x + 1\), assuming \(x eq 1\).

Step 5: Predict for k=4

Based on the pattern observed, we expect the simplified form of the expression to involve a polynomial of degree \(k-1\). For \(k=4\), we expect \(\frac{x^4 - 1}{x - 1}\) to simplify to \(x^3 + x^2 + x + 1\).

Key Concepts

Polynomials

Polynomials are an essential part of algebra that students will encounter frequently. They are expressions composed of variables and coefficients, combined using only addition, subtraction, and multiplication operations, and they do not involve division by a variable. For instance, a simple polynomial could be written as \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable.

One of the key characteristics of polynomials is their degree, which is determined by the highest power of the variable present in the expression. A polynomial of degree \( n \) is typically expressed in the form \( a_n x^n + a_{n-1}x^{n-1} + \, \ldots \, + a_1x + a_0 \).

In the context of the exercise, the polynomial \( x^k - 1 \) represents a difference of powers. The degree of this polynomial is \( k \), and understanding how to manipulate such polynomials through operations like factorization is crucial for solving algebraic problems.

Factorization

Factorization is the process of breaking down a complex expression into simpler components (factors) that, when multiplied together, yield the original expression. In algebra, this is particularly useful for simplifying expressions and solving equations.

Consider the polynomial \( x^k - 1 \) as seen in the exercise. It can be factored using the formula: \( x^k - 1 = (x-1)(x^{k-1} + x^{k-2} + \, \ldots \, + x + 1) \). This formula is a specific case of the difference of powers.

Here’s how it works visually:

• For \( k = 2 \), \( x^2 - 1 = (x-1)(x+1) \).
• For \( k = 3 \), \( x^3 - 1 = (x-1)(x^2 + x + 1) \).
• For \( k = 4 \), by following the pattern, \( x^4 - 1 = (x-1)(x^3 + x^2 + x + 1) \).

Factorization not only simplifies expressions but also reveals underlying patterns that can be exploited for mathematical insights and predictions.

Rational Expressions

Rational expressions are quotients of polynomials, similar to how rational numbers are fractions of integers. A rational expression is typically depicted in the form \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials, and \( Q(x) eq 0 \).

For the exercise presented, \( \frac{x^k - 1}{x - 1} \) is a rational expression. The denominator, \( x - 1 \), must not equal zero; thus \( x eq 1 \) ensures the expression remains valid. By performing factorization on the numerator and cancelling out common factors in the numerator and denominator, you can simplify the rational expression.

With each arithmetic manipulation, focusing on the restrictions imposed by domains (e.g., \( x = 1 \) being invalid here) is crucial. Simplifying rational expressions is a common task in algebra and helps in understanding the behavior of functions and in solving equations that involve polynomials.