Problem 34 - Beginning and Intermediate Algebra

Question

What binomial factors to a) \(\quad(y-9)(y+9) ?\) b) \(\quad(9-y)(9+y) ?\)

Step-by-Step Solution

Answer

a) The product of \((y-9)(y+9)\) is \(y^2 - 81\). b) The product of \((9-y)(9+y)\) is \(81 - y^2\).

Step 1: a) Finding the product of (y-9)(y+9)

To find the product, we can use the FOIL method (First, Outer, Inner, Last). Multiply the first terms, the outer terms, the inner terms, and the last terms, then sum the products: 1. First terms: \(y * y = y^2\) 2. Outer terms: \(y * 9 = 9y\) 3. Inner terms: \(-9 * y = -9y\) 4. Last terms: \(-9 * 9 = -81\) Now, combine the terms: \(y^2 + 9y - 9y - 81\) Since the +9y and -9y cancel out, the product is: \(y^2 - 81\)

Step 2: b) Finding the product of (9-y)(9+y)

Again, we will use the FOIL method (First, Outer, Inner, Last) to find the product: 1. First terms: \(9 * 9 = 81\) 2. Outer terms: \(9 * y = 9y\) 3. Inner terms: \(-y * 9 = -9y\) 4. Last terms: \(-y * y = -y^2\) Now, combine the terms: \(81 + 9y - 9y - y^2\) Since the +9y and -9y cancel out, the product is: \(81 - y^2\)

Key Concepts

FOIL Method

The FOIL Method is a technique used to multiply two binomials. It's a handy acronym that stands for First, Outer, Inner, and Last. This technique is particularly useful because it helps you break down the multiplication process into easy, manageable parts.

Here's how the FOIL Method works:

• First: Multiply the first terms in each binomial. For example, in a) \(y-9\)(y+9)\, these terms are \(y \times y = y^2\).
• Outer: Multiply the outer terms. In our example, this is \(y \times 9 = 9y\).
• Inner: Multiply the inner terms. For \(y-9\)(y+9)\, that would be \(-9 \times y = -9y\).
• Last: Multiply the last terms. Here, it's \(-9 \times 9 = -81\).

Finally, you gather all these products and combine like terms to simplify. For instance, in \( (y-9)(y+9)\), the middle terms cancel out (\(+9y-9y=0\)), leaving you with \(y^2 - 81\).

Using the FOIL method ensures you accurately multiply binomials, keeping track of all parts involved.

Difference of Squares

The Difference of Squares is a special algebraic pattern that appears when two squares are subtracted. It takes the form \(a^2 - b^2\).

This pattern is quite straightforward to factor using the formula:

• \(a^2 - b^2 = (a+b)(a-b)\)

Here's how it works with the exercise:

In \( (y-9)(y+9) = y^2 - 81 \), you have binomials that fit the \(a+b\) and \(a-b\) structure. Therefore, \(y^2 - 81\) can be seen as a difference of squares since \(y^2\) and \(81\) (which equals \(9^2\)) are perfect squares. Similarly, \( (9-y)(9+y) = 81 - y^2 \) also represents a difference of squares.

Recognizing this pattern helps simplify expressions quickly and easily, especially in tests or when solving equations.

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and arithmetic operations (like addition and multiplication). They are the foundation of algebra and appear in numerous forms.

Understanding algebraic expressions allows you to:

• Formulate mathematical models of real-world situations.
• Simplify and solve equations.
• Perform arithmetic operations on expressions.

When working with expressions like \(y^2 - 81\) or \(81 - y^2\), you're engaging with variables (\(y\)) and constants (like \(81\)). Knowing how to simplify and manipulate these types of expressions is essential in algebra.

For example, the solution to the exercise is more straightforward if you see \(y^2 - 81\) as \( (y-9)(y+9) \) from the difference of squares. This insight can reveal the factors and simplify the solving process. Algebraic expressions are versatile tools in mathematics.