Problem 46 - Precalculus: Mathematical for Calculus

Question

Assembling and Disassembling Partial Fractions The following expression is a partial fraction decomposition: $$ \frac{2}{x-1}+\frac{1}{(x-1)^{2}}+\frac{1}{x+1} $$ Use a common denominator to combine the terms into one fraction. Then use the techniques of this section to find its partial fraction decomposition. Did you get back the original expression?

Step-by-Step Solution

Answer

Yes, the original expression was recovered.

Step 1: Identify Common Denominator

Observe the denominators of each term: \((x-1)\), \((x-1)^2\), and \((x+1)\). The least common denominator (LCD), combining these, is \((x-1)^2(x+1)\).

Step 2: Rewrite Each Fraction

Express each term with the common denominator found:- \(\frac{2}{x-1}\) into \(\frac{2(x-1)(x+1)}{(x-1)^2(x+1)}\)- \(\frac{1}{(x-1)^2}\) remains \(\frac{1(x+1)}{(x-1)^2(x+1)}\)- \(\frac{1}{x+1}\) into \(\frac{1(x-1)^2}{(x-1)^2(x+1)}\).

Step 3: Combine Fractions Under the Common Denominator

Add the expressions derived in Step 2: \[\frac{2(x-1)(x+1) + (x+1) + (x-1)^2}{(x-1)^2(x+1)}\]

Step 4: Simplify the Numerator

Expand and combine like terms in the numerator: \[2(x^2-1) + x + 1 + (x^2 - 2x +1) = 2x^2 - 2 + x + 1 + x^2 - 2x +1\]Combine the terms: \[3x^2 - x\]

Step 5: Polynomial Long Division

Divide \(3x^2 - x\) by \((x-1)^2(x+1)\) to simplify the expression, confirming that the decomposition matches the original fractions.

Step 6: Compare with Original

The final expression after simplification can be decomposed as: \[\frac{2}{x-1} + \frac{1}{(x-1)^2} + \frac{1}{x+1}\]We find it matches the original expression given.

Key Concepts

Common Denominator

When combining fractions, it's essential to have a common denominator. Think of it as finding a shared baseline that all fractions agree upon. In the exercise, we had three denominators: \((x-1)\), \((x-1)^2\), and \((x+1)\). The least common denominator (LCD) is the smallest expression that each term's denominator can divide into without remainder.

The process to find the LCD involves looking at each unique factor in the denominators and taking the highest power of these factors. Here, the LCD is \((x-1)^2(x+1)\). Each original term's denominator can now be transformed by multiplying and adjusting the numerator accordingly, allowing them to combine easily.

Using a common denominator helps streamline calculations and removes complexity from the equation. Without it, we could not have easily added the fractions together.

Polynomial Long Division

Polynomial long division is a method similar to numerical division, but it's suited for variables and their exponents. Once we've combined our fractions over a common denominator, we might get a polynomial that needs further simplifying.

In the exercise, after simplifying the terms, we ended up with the polynomial \(3x^2 - x\). To confirm that our decomposed fractions were correct, we had to perform polynomial long division with \((x-1)^2(x+1)\) as the divisor.

Executing this division is about unfolding step by step:

• Compare the leading term of the polynomial with the leading term of the divisor.
• Multiply accordingly, subtract from the original, and bring down the next term.
• Repeat until the remainder's degree is less than that of the divisor.

Polynomial long division is crucial because it tells us if our decomposition or simplification is correct.

Rational Expressions

Rational expressions are fractions where both the numerator and the denominator are polynomials. They appear extensively in algebra, calculus, and real-world applications like physics and engineering.

This exercise is a great example of working with rational expressions. When faced with a rational expression, we often need to break it down into simpler "partial fractions" or combine fractions under a common denominator, as seen in the solution steps.

These processes revolve around:

• Understanding degrees: The degree of the numerator versus the denominator influences how we simplify.
• Simplification techniques like factoring or polynomial division.

It's essential to practice manipulating these expressions to achieve clarity and accuracy in solutions. Mastering rational expressions opens up deeper mathematical concepts and enhances problem-solving skills.