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Chapter 5: Problem 38

Find each product. $$ \left(7 x+\frac{3}{7}\right)\left(7 x-\frac{3}{7}\right) $$

Short Answer

Expert verified
The product is \( 49x^2 - \frac{9}{49} \).

Step by step solution

01

- Identify the expression formats

The given expression is \[ \left(7x + \frac{3}{7}\right) \left(7x - \frac{3}{7}\right) \] This is a product of two binomials in the format of \( (a+b)(a-b) \), which is known as the difference of squares formula.
02

- Apply the difference of squares formula

Recall that the difference of squares formula states \[ (a + b)(a - b) = a^2 - b^2 \] Here, \( a = 7x \) and \( b = \frac{3}{7} \).
03

- Calculate \( a^2 \)

\( a^2 \) means to square \(a\). So, \( (7x)^2 = 49x^2 \).
04

- Calculate \( b^2 \)

\( b^2 \) means to square \(b\). So, \( \left( \frac{3}{7} \right)^2 = \frac{9}{49} \).
05

- Substitute into the formula

Now substitute \( a^2 \) and \( b^2 \) back into the difference of squares formula: \[ (7x + \frac{3}{7})(7x - \frac{3}{7}) = 49x^2 - \frac{9}{49} \].

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Key Concepts

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

binomials
Binomials are algebraic expressions that contain two distinct terms. These terms can be variables, constants, or a combination of both. For example, in the exercise given, the expressions \(7x + \frac{3}{7}\) and \(7x - \frac{3}{7}\) are binomials.

Binomials are often combined using addition or subtraction. Here, they are part of a multiplication problem. Understanding how to handle binomials is key to solving many algebraic problems. The structure of binomials makes them suitable for certain algebraic formulas, such as the difference of squares.
multiplication of algebraic expressions
When multiplying algebraic expressions, you need to apply various algebraic rules. One common rule used in the exercise is the difference of squares formula: \((a + b)(a - b) = a^2 - b^2\).

In this case, the expressions \(7x + \frac{3}{7}\) and \(7x - \frac{3}{7}\) are multiplied together. By recognizing that they follow the difference of squares format, we can simplify the multiplication process considerably.

The general steps include:
  • Recognizing the format
  • Applying the appropriate algebraic formula
  • Simplifying the terms by performing operations like squaring each term
This method simplifies complex expressions into simpler forms, making them easier to work with.
squaring terms
Squaring a term means multiplying that term by itself. In the exercise, squaring is done for both \(7x\) and \(\frac{3}{7}\).

For example, when we square the term \(7x\): \((7x)^2 = 49x^2\). Similarly, squaring \(\frac{3}{7}\) yields \(\left( \frac{3}{7} \right)^2 = \frac{9}{49}\).

Squaring both terms individually allows us to apply the difference of squares formula efficiently and correctly. Understanding how to handle squares and square roots is a fundamental part of algebra that will come up in various tasks, making this knowledge invaluable.

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Most popular questions from this chapter

Find each product. \(-4 t(t+3)^{3}\)

Write each product as a sum of terms. Write answers with positive exponents only. Simplify each term. See Section 1.8 $$ \frac{1}{4 y}\left(y^{4}+6 y^{2}+8\right) $$

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Match each expression in Column I with the equivalent expression in Column II. Choices in Column II may be used once, more than once, or not at all. An Exercise \(17, x \neq 0 .\). I (a) \(-2^{-4}\) (b) \((-2)^{-4}\) (c) \(2^{-4}\) (d) \(\frac{1}{2^{-4}}\) (e) \(\frac{1}{-2^{-4}}\) (f) \(\frac{1}{(-2)^{-4}}\) II \(\mathbf{A} .8\) \(\mathbf{B} .16\) \(\mathbf{C} .-\frac{1}{16}\) \(\mathbf{D} .-8\) \(\mathbf{E} .-16\) \(\mathbf{F} . \frac{1}{16}\)
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