Problem 65
Question
Multiply. \(\left(x+\frac{1}{7}\right)\left(x-\frac{3}{7}\right)\)
Step-by-Step Solution
Verified Answer
The product is \(x^2 - \frac{9}{49}\).
1Step 1: Identify the expression type
The expression \((x+\frac{1}{7})(x-\frac{3}{7})\) is a product of two binomials of the form \((a+b)(a-b)\). This indicates that we can utilize the difference of squares formula.
2Step 2: Apply the difference of squares formula
For expressions of the form \((a+b)(a-b)\), the formula is \(a^2-b^2\). Here, \(a = x\) and \(b = \frac{3}{7}\). Using the formula, we have:\[a^2-b^2 = x^2 - \left(\frac{3}{7}\right)^2.\]
3Step 3: Simplify the expression
Calculate \(b^2\), where \(b = \frac{3}{7}\). Then, substitute back into the difference of squares formula:\[\left(\frac{3}{7}\right)^2 = \frac{9}{49},\]leading to the simplified expression:\[x^2 - \frac{9}{49}.\]
Key Concepts
Binomial MultiplicationAlgebraic ExpressionsSimplifying Expressions
Binomial Multiplication
When we talk about binomial multiplication, we're talking about multiplying two expressions that each have two terms. Binomials are a special type of polynomial. In simpler terms, they're any expression that you could write as \((a + b)\). This means they have exactly two terms.
To multiply binomials, you can use different methods, such as the FOIL method, which stands for First, Outer, Inner, and Last. This method multiplies each part of the first binomial with each part of the second binomial. But sometimes we can make our lives easier.
With specific types of binomials, such as in the exercise \((x+\frac{1}{7})(x-\frac{3}{7})\), you can use special patterns like the difference of squares. This pattern helps you quickly multiply without doing each tiny multiplication like with FOIL. You can jump straight to using a formula!
To multiply binomials, you can use different methods, such as the FOIL method, which stands for First, Outer, Inner, and Last. This method multiplies each part of the first binomial with each part of the second binomial. But sometimes we can make our lives easier.
With specific types of binomials, such as in the exercise \((x+\frac{1}{7})(x-\frac{3}{7})\), you can use special patterns like the difference of squares. This pattern helps you quickly multiply without doing each tiny multiplication like with FOIL. You can jump straight to using a formula!
Algebraic Expressions
Algebraic expressions are built using numbers, variables, and operation symbols like addition or subtraction. The expression may also include fractions, as we see in the given exercise. Understanding how to manipulate algebraic expressions is key to solving equations.
In the exercise \((x+\frac{1}{7})(x-\frac{3}{7})\), each binomial, or bracketed term, is an algebraic expression. Here, \(x\) is a variable representing a number, and \(\frac{1}{7}\) and \(-\frac{3}{7}\) are constants. These expressions show relationships between quantities.
Algebraic expressions can come in many forms. They may involve many variables, constants, and operations. The beauty of algebra is in its balance and precision, allowing us to describe complex ideas with simple symbols.
In the exercise \((x+\frac{1}{7})(x-\frac{3}{7})\), each binomial, or bracketed term, is an algebraic expression. Here, \(x\) is a variable representing a number, and \(\frac{1}{7}\) and \(-\frac{3}{7}\) are constants. These expressions show relationships between quantities.
Algebraic expressions can come in many forms. They may involve many variables, constants, and operations. The beauty of algebra is in its balance and precision, allowing us to describe complex ideas with simple symbols.
Simplifying Expressions
Simplifying expressions makes them easier to understand and work with. It's about paring down an expression to its simplest form so it is easier to read at a glance.
In the practice problem, after applying the difference of squares formula, we reached the expression: \(x^2 - \frac{9}{49}\). To simplify, you calculate \(b^2\), which for \(b = \frac{3}{7}\), is \(\left(\frac{3}{7}\right)^2 = \frac{9}{49}\). Thus the expression \(x^2 - \left(\frac{3}{7}\right)^2\) simplifies to \(x^2 - \frac{9}{49}\).
Simplifying often involves applying identities like the difference of squares or cancelling terms. The goal is always to make your math expression look its best, free of unnecessary terms or factors.
In the practice problem, after applying the difference of squares formula, we reached the expression: \(x^2 - \frac{9}{49}\). To simplify, you calculate \(b^2\), which for \(b = \frac{3}{7}\), is \(\left(\frac{3}{7}\right)^2 = \frac{9}{49}\). Thus the expression \(x^2 - \left(\frac{3}{7}\right)^2\) simplifies to \(x^2 - \frac{9}{49}\).
Simplifying often involves applying identities like the difference of squares or cancelling terms. The goal is always to make your math expression look its best, free of unnecessary terms or factors.
Other exercises in this chapter
Problem 65
Simplify each expression. Write each result using positive exponents only. $$ \frac{\left(x y^{3}\right)^{5}}{(x y)^{-4}} $$
View solution Problem 65
Mixed Practice Multiply. $$ (3 a+1)^{2} $$
View solution Problem 65
Simplify each expression. $$ (2 x)^{0} $$
View solution Problem 66
Write each polynomial in descending powers of the variable and with no missing powers. See Example 15. $$ 11 z+4 z^{4} $$
View solution