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Chapter 6: Problem 186

Factor completely using the difference of squares pattern, if possible. $$ 36 p^{2}-49 q^{2} $$

### Short Answer

Expert verified

(6p + 7q)(6p - 7q) .

## Step by step solution

01

## Identify terms

Observe the expression: [36 p^{2} - 49 q^{2} ]. Identify each term as a perfect square.

02

## Express as squares

Write each term as a square: [36 p^{2} = (6p)^2 ] and [49 q^{2} = (7q)^2 ].

03

## Apply difference of squares formula

Use the difference of squares formula: [a^2 - b^2 = (a + b)(a - b) ]. Here, [a = 6p ] and [b = 7q ].

04

## Substitute and simplify

Substitute [a = 6p ] and [b = 7q ] into the formula to get: [(6p + 7q)(6p - 7q) ].

## 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 powerful tool in algebra for factorizing certain kinds of expressions. It applies to expressions that can be written in the form \[ a^2 - b^2 \].

The key to using this method is recognizing whether both terms in the expression are perfect squares.

Once identified, we can use the formula: \[ a^2 - b^2 = (a + b)(a - b) \].

In the given exercise, we have \[ 36p^2 - 49q^2 \]. Notice that both 36 and 49 are perfect squares, and so are the variables \[ p^2 \] and \[ q^2 \].

Rewriting, we find that \[ 36p^2 = (6p)^2 \] and \[ 49q^2 = (7q)^2 \].

Thus, \[ 36p^2 - 49q^2 \] can be transformed into \[ (6p)^2 - (7q)^2 \]. Using the difference of squares formula, we get \[ (6p + 7q)(6p - 7q) \].

###### Perfect Square

A perfect square is any number or expression that can be written as the square of another number or expression.

For instance, 36 is a perfect square because it can be written as \[ 6^2 \]. Similarly, \[ p^2 \] is a perfect square because it is \[ (p)^2 \].

Identifying perfect squares is crucial when dealing with algebraic expressions and factorization problems, particularly those involving the difference of squares.

When you encounter an expression such as \[ a^2 - b^2 \], check if \[ a^2 \] and \[ b^2 \] represent perfect squares.

In our exercise, \[ 36p^2 \] is a perfect square as it equals \[ (6p)^2 \], and similarly, \[ 49q^2 \] is a perfect square because it equals \[ (7q)^2 \].

Recognizing perfect squares can simplify complex algebraic expressions and aid in various factorization techniques.

###### Algebraic Expressions

Algebraic expressions consist of numbers, variables, and operations. They form the foundation of algebra and are used to represent mathematical relationships and problems.

For example, in the expression \[ 36p^2 - 49q^2 \], we deal with terms involving both numbers (coefficients) and variables.

To simplify or factorize such expressions, it helps to understand the properties of numbers and algebraic rules.

In our exercise, the expression is \[ 36p^2 - 49q^2 \]. This can be factorized using the difference of squares method since both terms are perfect squares: \[ (6p)^2 - (7q)^2 \].

Applying the difference of squares formula yields \[ (6p + 7q)(6p - 7q) \].

Understanding how to manipulate and factorize algebraic expressions is an essential skill in solving mathematical problems efficiently.

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