Problem 60 - Algebra and Trigonometry

Question

Multiply the algebraic expressions using a Special Product Formula, and simplify. \((y-3)(y+3)\)

Step-by-Step Solution

Answer

The simplified expression is \(y^2 - 9\).

Step 1: Identify the Special Product Formula

The expression \((y-3)(y+3)\) fits the form of the difference of squares, which is \((a-b)(a+b) = a^2-b^2\). In this case, \(a = y\) and \(b = 3\).

Step 2: Apply the Difference of Squares Formula

Using the formula, calculate \(a^2-b^2\). Substitute \(a = y\) and \(b = 3\) to obtain \(y^2 - 3^2\).

Step 3: Simplify the Expression

Simplify \(y^2 - 3^2\) to get \(y^2 - 9\). This is the simplified form of the expression.

Key Concepts

Difference of Squares

In algebra, the difference of squares is a special product pattern that simplifies the multiplication of certain algebraic expressions. The difference of squares formula is written as \((a-b)(a+b) = a^2-b^2\). Here, we are looking at the product of two binomials: one with a subtraction and one with an addition between the same two terms.

It can be seen commonly in expressions like \((y-3)(y+3)\). To identify this pattern:

• Look for two terms that are the same in both binomials.
• Check if one binomial is an addition and the other a subtraction between the same numbers.

When you see this arrangement, you can directly apply the difference of squares formula to simplify the expression quickly.

The key advantage of recognizing the difference of squares is that it allows for immediate simplification without requiring detailed multiplication of each term.

Simplifying Algebraic Expressions

Simplifying algebraic expressions is a crucial skill in mathematics. It involves reducing an expression to its simplest form. When you have a pattern like the difference of squares, simplifying becomes much easier.

In our exercise, applying the difference of squares, results in an expression \(y^2 - 9\). This is because, when simplifying:

• First, calculate the square of the terms involved: \(y^2\) and \(-3^2\).
• Subtract the second square from the first: \(y^2 - 9\).

This concise form \(y^2 - 9\) is more elegant and easier to work with for further algebraic manipulations, or when solving for values of \(y\). Simplifying expressions helps in revealing more straightforward relations that might be hidden otherwise.

Algebraic Multiplication

Algebraic multiplication involves applying arithmetic rules to multiply algebraic terms or expressions. Knowing special product formulas, like the difference of squares, helps to deal with algebraic multiplication efficiently.

For the problem \((y-3)(y+3)\), instead of multiplying each term individually, we recognize it as a special product. Here's how understanding algebraic multiplication here helps:

• It saves time by leveraging known formulas.
• Reduces potential errors by simplifying operations.

In algebraic multiplication, identifying such patterns as the difference of squares simplifies calculations considerably. Thus, it is not only about knowing the rules but also spotting opportunities to apply them for efficiency and clarity in problem-solving.