Problem 13 - Elementary Algebra

Question

Factor completely. $$ 4 y 2-81 $$

Step-by-Step Solution

Answer

The complete factorization is \((2y - 9)(2y + 9)\).

Step 1: Identify the difference of squares formula

The expression is in the form of a difference of squares. Recall that the difference of squares formula is given by \( a^2 - b^2 = (a-b)(a+b) \). This formula will be useful to factor the expression.

Step 2: Recognize terms as squares

Notice that \( 4y^2 - 81 \) can be rewritten as \( (2y)^2 - 9^2 \). Here, \( (2y)^2 \) is a perfect square and \( 9^2 \) is also a perfect square.

Step 3: Apply the difference of squares formula

Using the difference of squares formula, replace \( a \) with \( 2y \) and \( b \) with \( 9 \). Thus, the expression becomes \( (2y - 9)(2y + 9) \).

Step 4: Verify your factorization

Expand the factors to check if they result in the original expression. \( (2y - 9)(2y + 9) = 2y \cdot 2y + 2y \cdot 9 - 9 \cdot 2y - 9 \cdot 9 = 4y^2 - 81 \). The factorization is correct.

Key Concepts

Difference of Squares

The difference of squares is a fundamental concept in algebra that facilitates the factorization of certain types of expressions. It is given by the formula

• \( a^2 - b^2 = (a-b)(a+b) \)

This formula shows how a subtraction between two perfect squares can be expressed as a product of two binomials. This pattern is immensely helpful in simplifying polynomial expressions. In the expression \( 4y^2 - 81 \), recognizing each term as a perfect square allows us to apply this formula effectively.

Notice that \( 4y^2 \) is the square of \( 2y \), and \( 81 \) is the square of \( 9 \). By setting \( a = 2y \) and \( b = 9 \), it becomes easier to see that the expression fits the difference of squares form. This realization allows us to confidently rewrite \( 4y^2 - 81 \) as \( (2y-9)(2y+9) \), simplifying the original expression.

Polynomials

Polynomials are algebraic expressions consisting of variables and coefficients, which are combined using only addition, subtraction, multiplication, and non-negative integer exponents. They are a core part of algebra and appear in various formats.

In polynomial expressions, each single term is a monomial, and combining such terms leads to polynomials. The given expression, \( 4y^2 - 81 \), is a simple polynomial with two terms. It highlights the concept of subtracting squares, emphasizing its straightforward nature.

The factorization of polynomials such as this one heavily relies on identifying patterns like the difference of squares. Knowing how to factor and manipulate polynomials ensures you can solve larger and more complex algebraic problems efficiently.

Algebraic Expressions

Algebraic expressions are mathematical phrases involving variables, numbers, and operations. They form the building blocks of algebraic equations and functions.

• Examples of operations include addition, subtraction, multiplication, and division.
• Variables can represent unknown values, making algebraic expressions flexible and powerful for problem-solving.

The expression \( 4y^2 - 81 \) is an algebraic expression that can also be seen as a polynomial characterized by its two terms. Recognizing patterns within algebraic expressions helps identify how best to simplify or manipulate them.

This understanding aids in making expressions like \( 4y^2 - 81 \) more manageable by converting them into their factored forms. By accomplishing this, computations can be simplified, solutions can be found more easily, and the properties of the expression become more apparent.