Problem 64 - Algebra and Trigonometry

Question

Perform the addition or subtraction and simplify. $$ \frac{1}{x+1}-\frac{2}{(x+1)^{2}}+\frac{3}{x^{2}-1} $$

Step-by-Step Solution

Answer

The simplified expression is \( \frac{x^2 + x + 4}{(x+1)^2(x-1)} \).

Step 1: Rewrite the Expression with a Common Denominator

First, find a common denominator for all three fractions. The denominators are \((x+1)\), \((x+1)^2\), and \(x^2-1\). Note that \(x^2-1\) can be factored as \((x-1)(x+1)\). The least common multiple of these denominators is \((x+1)^2(x-1)\). Rewrite each fraction with this common denominator.

Step 2: Convert First Fraction

Rewrite \( \frac{1}{x+1} \) with the common denominator. Multiply numerator and denominator by \((x+1)(x-1)\):\[ \frac{1}{x+1} = \frac{1 \cdot (x+1)(x-1)}{(x+1)^2(x-1)} = \frac{(x+1)(x-1)}{(x+1)^2(x-1)} \]

Step 3: Convert Second Fraction

Rewrite \( \frac{2}{(x+1)^2} \) with the common denominator. Multiply numerator and denominator by \((x-1)\):\[ \frac{2}{(x+1)^2} = \frac{2 \cdot (x-1)}{(x+1)^2(x-1)} = \frac{2(x-1)}{(x+1)^2(x-1)} \]

Step 4: Convert Third Fraction

Rewrite \( \frac{3}{x^2-1} \) with the common denominator. Multiply numerator and denominator by \((x+1)\):\[ \frac{3}{x^2-1} = \frac{3 \cdot (x+1)}{(x+1)^2(x-1)} = \frac{3(x+1)}{(x+1)^2(x-1)} \]

Step 5: Combine the Fractions

Now that all fractions have a common denominator, combine them:\[ \frac{(x+1)(x-1) - 2(x-1) + 3(x+1)}{(x+1)^2(x-1)} \]

Step 6: Expand and Simplify the Numerator

Expand and simplify the numerator:\[ (x+1)(x-1) = x^2-1 \quad \text{and} \quad 2(x-1) = 2x-2 \quad \text{and} \quad 3(x+1) = 3x+3 \]Combine these to get:\[ x^2 - 1 - 2x + 2 + 3x + 3 = x^2 + x + 4 \]

Step 7: Simplify the Entire Expression

The combined fraction becomes:\[ \frac{x^2 + x + 4}{(x+1)^2(x-1)} \]No further simplification is possible.

Key Concepts

Addition of Rational Expressions

When adding rational expressions, it's crucial to ensure they have the same denominator. Just like with numerical fractions, you can't simply add the numerators without first aligning the denominators. If their denominators differ, you'll need to find a common denominator. For the given problem, the expressions we need to add and subtract are:

• \( \frac{1}{x+1} \)
• \( \frac{2}{(x+1)^2} \)
• \( \frac{3}{x^2-1} \)

To do this effectively, we must first express each fraction with this common denominator, allowing us to combine them just like we would with simple fractions. This step is vital and sets the stage for simplifying the overall expression.

Subtraction of Rational Expressions

Subtraction of rational expressions follows the same principles as addition, where having a common denominator is a prerequisite. In our exercise, we initially subtract \( \frac{2}{(x+1)^2} \) from \( \frac{1}{x+1} \). Before performing the subtraction, both rational expressions need to have identical denominators. Once a common denominator is established, as in this case with \((x+1)^2(x-1)\), we simply rearrange and modify the numerators accordingly. Subtraction often involves careful handling of signs, ensuring terms are subtracted correctly after rewriting the numerators. This requires close attention to ensure the entire numerical structure aligns correctly for simplification. The process of subtraction, when properly executed, simplifies the complexity of multiple rational expressions into a singular, more manageable form.

Common Denominators

A common denominator is a shared multiple of the original denominators in rational expressions. In our problem, the denominators involved are \((x+1)\), \((x+1)^2\), and \(x^2-1\). To make the expressions compatible for addition or subtraction, they must be rewritten with a shared denominator.Finding a common denominator can involve:

• Factoring each denominator, if necessary.
• Determining the least common multiple (LCM) of these factors.

In the example exercise, \(x^2-1\) is factored into \((x-1)(x+1)\). The resulting LCM, and thus our common denominator, is \((x+1)^2(x-1)\). This step ensures that each rational expression is rewritten in a compatible format, streamlining the process of performing arithmetic operations.

Polynomial Factoring

Polynomial factoring is a process that simplifies complex expressions, making them easier to handle in algebraic operations. In rational expressions like our given problem, factoring is used to break down polynomials in the denominators, as seen with \(x^2-1\), which factors to \((x-1)(x+1)\).The key benefits of polynomial factoring include:

• Revealing the roots of the polynomial.
• Simplifying the expression, which assists in finding common denominators.
• Highlighting opportunities for further simplification and manipulation.

In our exercise, once we factored \(x^2-1\), it allowed us to find a common denominator efficiently. Factoring transforms seemingly intricate algebraic expressions into more straightforward components, enhancing solvability for addition and subtraction tasks.