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Chapter 6: Problem 65

Write each rational expression in lowest terms. $$ \frac{a^{2}-b^{2}}{a^{2}+b^{2}} $$

Short Answer

Expert verified
\( \frac{(a - b)(a + b)}{a^{2}+b^{2}} \)

Step by step solution

01

Identify the expression

The given rational expression is \ \( \frac{a^{2}-b^{2}}{a^{2}+b^{2}} \).
02

Factorize the numerator

Notice that the numerator \( a^2 - b^2 \) is a difference of squares. It can be factorized using the difference of squares formula: \ \( a^2 - b^2 = (a - b)(a + b) \). So, \( \frac{a^{2}-b^{2}}{a^{2}+b^{2}} \) becomes \ \( \frac{(a - b)(a + b)}{a^{2}+b^{2}} \).
03

Simplify the expression

Observe if any common factors exist in the numerator and the denominator that can be canceled. Here, \( a^2 + b^2 \) in the denominator cannot be factorized further and is not a common factor with any part of the numerator. Hence, the expression in its lowest terms remains \ \( \frac{(a - b)(a + b)}{a^{2}+b^{2}} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Difference of Squares
The difference of squares is a key algebraic concept. It allows us to factorize expressions in the form of \( a^2 - b^2 \). This specific form can be rewritten using the special factorization formula:\[ a^2 - b^2 = (a - b)(a + b) \]This formula exploits the fact that the product of two conjugate binomials results in the difference of their squared terms. Using this method, intricate expressions are simplified for easier manipulation. Recognizing a difference of squares quickly is crucial.
For example:
  • \( x^2 - 9 \) becomes \( (x - 3)(x + 3) \)
  • \( 25y^2 - 16 \) becomes \( (5y - 4)(5y + 4) \)
Always remember that both terms must be perfect squares, with a subtraction sign between them. This fundamental principle makes factorization simpler in many algebra problems.
Factorization
Factorization entails breaking down an expression into its multiplicative components. It's like finding the building blocks of the given algebraic structure. To factorize means to express an algebraic formula as a product of its factors. There are numerous techniques for factorizing different kinds of expressions:
  • Common Factor Extraction: Taking out the greatest common factor from all terms.
  • Difference of Squares: Using the formula \( a^2 - b^2 = (a - b)(a + b) \).
  • Quadratic Trinomials: Factoring expressions like \( ax^2 + bx + c \).
In the exercise provided, we factorized the numerator \( a^2 - b^2 \) using the difference of squares method.
Resulting in:
\( \frac{(a - b)(a + b)}{a^2 + b^2} \)
This step rearranges the expression for easier cancellation of common factors, leading to simpler forms for solving or further manipulation.
Lowest Terms
Simplifying an expression to its lowest terms means reducing it to the simplest form possible, removing all common factors in the numerator and the denominator. Here are the steps:
  • Identify any common factors in the numerator and denominator.
  • Cancel out these common factors.
  • Ensure the expression cannot be simplified further.

In our example, we simplified \( \frac{a^2 - b^2}{a^2 + b^2} \). After factorization, we got:
\( \frac{(a - b)(a + b)}{a^2 + b^2} \)
Since \( a^2 + b^2 \) cannot be further factorized and doesn't share common factors with the numerator, this is the expression's lowest term.
Simplifying makes expressions more manageable and solvable, a crucial algebra skill.

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Most popular questions from this chapter

Multiply or divide as indicated. $$ \frac{(2 x+3)(x-4)}{(x+8)(x-4)} \div \frac{(x-4)(x+2)}{(x-4)(x+8)} $$

Write each rational expression in lowest terms. $$ \frac{p^{2}+q^{2}}{p^{2}-q^{2}} $$

Solve each problem. The maximum load of a horizontal beam that is supported at both ends varies directly as the width and the square of the height and inversely as the length between the supports. A beam \(6 \mathrm{~m}\) long, \(0.1 \mathrm{~m}\) wide, and \(0.06 \mathrm{~m}\) high supports a load of \(360 \mathrm{~kg}\). What is the maximum load supported by a beam \(16 \mathrm{~m}\) long, \(0.2 \mathrm{~m}\) wide, and \(0.08 \mathrm{~m}\) high?

Add or subtract as indicated. $$\frac{5}{x+2}+\frac{2}{x^{2}-2 x+4}-\frac{60}{x^{3}+8}$$

Multiply or divide as indicated. $$ \left(\frac{6 k^{2}-13 k-5}{k^{2}+7 k} \div \frac{2 k-5}{k^{3}+6 k^{2}-7 k}\right) \cdot \frac{k^{2}-5 k+6}{3 k^{2}-8 k-3} $$
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