Problem 16
Question
Multiply and reduce. Do some by calculator. $$\frac{c}{x^{2}-y^{2}} \cdot \frac{d}{x^{2}-y^{2}}$$
Step-by-Step Solution
Verified Answer
\(\frac{cd}{(x^2-y^2)^2}\)
1Step 1: Multiplying the fractions
To multiply two fractions, you multiply the numerators (the top numbers) together and the denominators (the bottom numbers) together. For these two fractions, multiply the numerators, which are just the variables c and d, and multiply the denominators, which both are the difference of squares, \(x^2-y^2\).
2Step 2: Apply the multiplication
After multiplying the numerators and denominators respectively, the new numerator is \(cd\) and the new denominator is \(x^2-y^2)^2\). This is because when you multiply expressions with the same base, you add the exponents.
3Step 3: Writing the simplified expression
The simplified expression after multiplying is \[\frac{cd}{(x^2-y^2)^2}\]. No further simplification can be done unless more information is provided about the variables c, d, x, and y.
Key Concepts
Difference of SquaresSimplifying Algebraic ExpressionsExponents in Algebra
Difference of Squares
When we come across the term difference of squares, it refers to a specific algebraic pattern that is an essential concept in simplifying expressions and solving equations. This pattern can be recognized when you have an expression of the form \( a^2 - b^2 \), where \( a \) and \( b \) are any algebraic expressions. It's called difference because it involves subtraction, and squares because each term is a square of something.
The power of recognizing this pattern comes from the fact that a difference of squares can always be factored into \( (a + b)(a - b) \) showing the relationship between the algebraic factors. For instance, in the exercise given, \( x^2 - y^2 \) is a difference of squares and can be factored into \( (x + y)(x - y) \) which sometimes aids in further simplification or solving equations.
Understanding how to recognize and factor the difference of squares is a stepping stone for tackling more complex algebraic expressions and is a frequent presence in problems involving multiplication and division of algebraic fractions.
The power of recognizing this pattern comes from the fact that a difference of squares can always be factored into \( (a + b)(a - b) \) showing the relationship between the algebraic factors. For instance, in the exercise given, \( x^2 - y^2 \) is a difference of squares and can be factored into \( (x + y)(x - y) \) which sometimes aids in further simplification or solving equations.
Understanding how to recognize and factor the difference of squares is a stepping stone for tackling more complex algebraic expressions and is a frequent presence in problems involving multiplication and division of algebraic fractions.
Simplifying Algebraic Expressions
In algebra, the process of simplifying is key to making expressions easier to work with. To simplify an algebraic expression, one must combine like terms, factor when possible, and reduce fractions to their lowest terms.
When an expression involves fractions, as seen in the exercise with \( \frac{cd}{(x^2-y^2)^2} \) there is often a misconception that simplification always involves making the expression shorter. However, this isn't always the case. To truly simplify means to put the expression into its most manageable form, which sometimes may look longer than the original. In simplifying the expression from the exercise, given that no further information is available about the variables, we leave it as it is because it's already in its simplest form. Simplification is essential as it aids in the evaluation of expressions, solving equations, and may reduce errors in computation.
When an expression involves fractions, as seen in the exercise with \( \frac{cd}{(x^2-y^2)^2} \) there is often a misconception that simplification always involves making the expression shorter. However, this isn't always the case. To truly simplify means to put the expression into its most manageable form, which sometimes may look longer than the original. In simplifying the expression from the exercise, given that no further information is available about the variables, we leave it as it is because it's already in its simplest form. Simplification is essential as it aids in the evaluation of expressions, solving equations, and may reduce errors in computation.
Exponents in Algebra
The power of exponents is one of the fundamental concepts in algebra. Exponents indicate how many times a number or an expression is multiplied by itself. In the exercise, when we multiply the denominators and encounter expressions such as \( (x^2)^2 \) we adhere to the law of exponents that stipulates: to multiply powers with the same base, you add their exponents. This results in \( x^{2 + 2} \) or \( x^4 \) in our instance.
Understanding how to manipulate and simplify expressions with exponents is crucial for higher-level math, including algebraic expressions involving fractions. Treat exponents carefully, especially in algebraic fractions, to avoid common mistakes such as incorrect distribution or misinterpreting the order of operations. Mastery of the rules governing exponents facilitates the simplification process and prepares students for complexities that arise in higher mathematics.
Understanding how to manipulate and simplify expressions with exponents is crucial for higher-level math, including algebraic expressions involving fractions. Treat exponents carefully, especially in algebraic fractions, to avoid common mistakes such as incorrect distribution or misinterpreting the order of operations. Mastery of the rules governing exponents facilitates the simplification process and prepares students for complexities that arise in higher mathematics.
Other exercises in this chapter
Problem 15
Reduce to lowest terms. Write your answers without negative exponents. Do some algebraic fractions by calculator. $$\frac{2 a b}{6 b}$$
View solution Problem 15
$$4 x^{2} y+c x y^{2}+3 x y^{3}$$
View solution Problem 16
Solve for \(x\). Assume the integers in these equations to be exact numbers, and leave your answers in fractional form. \(\frac{x}{4}+\frac{x}{6}+\frac{x}{8}=26
View solution Problem 16
Combine and simplify. Try some by calculator. $$\frac{3}{x}+\frac{2}{x}-\frac{1}{x}$$
View solution