Problem 36 - College Algebra

Question

In Exercises \(31-40,\) factor the difference of two squares. $$36 x^{2}-49 y^{2}$$

Step-by-Step Solution

Answer

The factored form of \(36x^2 - 49y^2\) is \((6x + 7y)(6x - 7y)\)

Step 1: Determine the square roots

We start by finding the square roots of the two terms. The square root of \(36x^2\) is \(6x\) and the square root of \(49y^2\) is \(7y\). Therefore, \(a = 6x\) and \(b = 7y\). Now we have the necessary elements to apply the formula.

Step 2: Apply the difference of squares formula

Next, we apply the formula for the difference of squares. For the given expression \(a^2 - b^2\), the formula is \((a+b)(a-b)\). In our problem, \(a\) equals \(6x\) and \(b\) equals \(7y\), so the difference of squares is \((6x + 7y)(6x - 7y)\).

Key Concepts

Difference of Squares

The difference of squares is a special algebraic expression that consists of two squared terms subtracted from one another. It looks like this: \(a^2 - b^2\). This expression can be factored using the formula \((a + b)(a - b)\). The formula is derived from multiplying the binomials and observing that the middle terms cancel each other out. Hence, the result is the original difference of squares.

• Key Properties: Both terms must be perfect squares.
• Requirement: Subtraction (difference) between the terms is crucial.

Once you have identified the squared terms and verified they meet the requirements, you confidently apply the formula to simplify expressions, like in the exercise given. This method is powerful for simplifying complex equations and useful in various algebraic applications.

Polynomial

A polynomial is an algebraic expression made up of variables and constants combined using addition, subtraction, multiplication, and non-negative integer exponents of variables. Each part of a polynomial separated by addition or subtraction is called a term. For example, in the expression \(36x^2 - 49y^2\), there are two terms: \(36x^2\) and \(-49y^2\).

Polynomials can be classified by:

• Degree: The highest power of the variable in the polynomial (like "2" in \(x^2\)).
• Number of terms: One (monomial), two (binomial), or more (trinomial and beyond).

This classification helps in understanding and factoring, as different kinds of polynomials may require different factoring techniques. Recognizing our example as a binomial polynomial is essential when using the difference of squares method.

Algebraic Expressions

An algebraic expression is a mathematical phrase that can contain numbers, variables (like \(x\) and \(y\)), and operators (such as +, -, *, /). These expressions can represent real-world quantities and are foundational in algebra, which helps solve equations and model situations.

Every algebraic expression can be simplified or transformed using various algebraic techniques, such as:

• Simplifying expression by combining like terms.
• Factoring, such as identifying a difference of squares.

In the context of the exercise, \(36x^2 - 49y^2\) is the given algebraic expression. Understanding how to manipulate such expressions using algebraic rules is essential for solving problems in algebra. By following logical steps, you can transform complex expressions into simplified and useful forms.

Square Root

The square root is a fundamental concept in mathematics, representing a value that, when multiplied by itself, gives the original number. In simpler terms, if \(b\) is the square root of \(a\), then \(b^2 = a\).

• Symbol: The square root is represented by the radical sign \(\sqrt{}\).
• Key Example: For \(36x^2\), the square root is \(6x\) because \((6x)^2 = 36x^2\).

Finding the square roots of terms is crucial in applying the difference of squares formula. By accurately determining \(a\) and \(b\) in the exercise example \(36x^2 - 49y^2\), you enable the effective use of the \((a+b)(a-b)\) formula. This ability to find square roots equips students to tackle a variety of algebraic problems.