Problem 54 - College Algebra

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: Simplify the Expression for k=1

For \(k=1\), substitute \(k\) in the expression \(\frac{x^{k}-1}{x-1}\). This gives us \(\frac{x^{1}-1}{x-1} = \frac{x-1}{x-1}\). Simplifying this will give us \(1\), since \(x-1\) cancels out.

Step 2: Simplify the Expression for k=2

For \(k=2\), substitute \(k\) in the expression \(\frac{x^{k}-1}{x-1}\). This becomes \(\frac{x^{2}-1}{x-1}\). Recognize that \(x^2 - 1\) can be factored as \((x-1)(x+1)\). We have \(\frac{(x-1)(x+1)}{x-1}\). Simplifying gives us \(x+1\), since \(x-1\) cancels out.

Step 3: Simplify the Expression for k=3

For \(k=3\), substitute \(k\) in the expression \(\frac{x^{k}-1}{x-1}\). This becomes \(\frac{x^{3}-1}{x-1}\). Recognize that \(x^3 - 1\) can be factored as \((x-1)(x^2+x+1)\). We have \(\frac{(x-1)(x^2+x+1)}{x-1}\). Simplifying gives us \(x^2 + x + 1\), since \(x-1\) cancels out.

Step 4: Predict the Result for k=4

Based on the previous steps, we see a pattern where \(k=1\) gives \(1\), \(k=2\) gives \(x+1\), and \(k=3\) gives \(x^2+x+1\). Each subsequent result is building upon powers of \(x\): \ - \(k=1\): \(x^0\) term is 1 \ - \(k=2\): \(x^1 + x^0\) \ - \(k=3\): \(x^2 + x^1 + x^0\) \ Following this pattern, for \(k=4\), we expect the result to be \(x^3 + x^2 + x + 1\) as it follows the form \(x^{k-1} + x^{k-2} + ... + x^0\).

Key Concepts

Polynomial Division

Polynomial Division is a process used to simplify expressions by dividing polynomials. It is similar to long division with numbers, but instead, we deal with the variables and coefficients in polynomial expressions. In this exercise, polynomial division is applied to expressions like \( \frac{x^{k}-1}{x-1} \). This form is simplified by dividing the polynomial in the numerator, \( x^k - 1 \), by the polynomial in the denominator, \( x-1 \).

• When \( k = 1 \), the expression simplifies instantly, with terms canceling out, resulting in a value of \( 1 \).
• For higher values of \( k \), factoring is first applied to the numerator before polynomial division simplifies the expression.
• The concept follows systematically for different powers of \( k \) and shows how lower-order terms contribute to simplifying polynomial expressions.

Understanding polynomial division fundamentally enhances your ability to manipulate and simplify complex algebraic expressions efficiently. It is widely applicable when solving mathematical problems involving polynomial expressions.

Factoring Polynomials

Factoring Polynomials breaks down complicated expressions into simpler components or factors. In algebra, recognizing and applying patterns to factor polynomials like \( x^k - 1 \) is crucial.

• For \( k = 2 \), \( x^2 - 1 \) is factored into \((x - 1)(x + 1)\) using the difference of squares.
• For \( k = 3 \), \( x^3 - 1 \) is factored as \((x - 1)(x^2 + x + 1)\).
• As \( k \) increases, these polynomials build into more complex products that can be simplified effectively using their factors.

Factoring provides critical insight into the structure of polynomial expressions, making it easier to solve equations and understand their behavior. It also reduces complicated expressions, paving the way for easier calculation, particularly when dividing polynomials.

Mathematical Patterns

Mathematical patterns play a vital role in predicting the outcome of algebraic expressions when dealing with sequences. Recognizing patterns helps simplify complex processes by predicting outcomes based on previous results.

• In this exercise, we observe that each result when simplified follows a straightforward pattern: \( x^{k-1} + x^{k-2} + ... + x^0 \).
• Knowing the terms of these patterns for past \( k \) values allows us to deduce that the result for \( k = 4 \) would naturally extend to \( x^3 + x^2 + x + 1 \).
• This predictable progression is critical when solving algebraic expressions without recalculating every new term individually.

Recognizing and applying these mathematical patterns can empower students to handle algebraic problems more confidently and predict much larger sequences with ease.

Sequence Prediction

Sequence Prediction refers to identifying the next logical element in an ordered sequence of numbers or expressions. By understanding and analyzing previous results, we can foresee upcoming sequence terms.

• Starting from the simple sequence of results for each \( k \), we're looking at patterns developed in polynomial simplification.
• This exercise shows such patterns for \( k = 1, 2, 3 \), and by understanding the pattern, we predict the result for \( k = 4 \) to be \( x^3 + x^2 + x + 1 \).
• The sequence is formed by adding terms of progressively lower degree of \( x \), creating a direct analogy with known patterns like the sum of geometric progressions.

Mastering sequence prediction requires practice and improves analytical skills by encouraging the understanding of relationships between mathematical expressions and successive terms. It is an essential skill for algebra and further mathematical studies.