Problem 35 - Algebra for College Students

Question

Find each indicated product. Remember the shortcut for multiplying binomials and the other special patterns we discussed in this section. $$(3 y-1)(3 y+1)$$

Step-by-Step Solution

Answer

The product is \(9y^2 - 1\).

Step 1: Identify the Formula

Recognize that the given expression \((3y - 1)(3y + 1)\) fits the form of a difference of squares, \((a - b)(a + b) = a^2 - b^2\). Here, \(a = 3y\) and \(b = 1\).

Step 2: Apply the Pattern

Using the difference of squares pattern \((a - b)(a + b) = a^2 - b^2\), substitute \(a = 3y\) and \(b = 1\) into the equation: \((3y)^2 - (1)^2\).

Step 3: Square the Terms

Calculate \((3y)^2\) and \(1^2\) separately. \((3y)^2 = 9y^2\) and \((1)^2 = 1\).

Step 4: Subtract to Find the Product

Subtract \(1\) from \(9y^2\): \(9y^2 - 1\).

Key Concepts

Difference of Squares

The difference of squares is a very useful algebraic pattern. It allows us to factor certain types of polynomials quickly. This pattern is expressed as

• \[(a - b)(a + b) = a^2 - b^2\]

Here, \(a\) and \(b\) are any algebraic expressions. When you multiply a binomial pair that fits this pattern, your result will always be a difference of two squares. This is because:

• (a - b) means there's a subtraction (\(-b\)).
• (a + b) means there's an addition (\(+b\)).
• When these are multiplied together, the sum (\(+\)) and difference (\(-\)) appearances cause the middle terms to cancel each other out.

In our example, \((3y - 1)(3y + 1)\), notices how:

• \(a = 3y\)
• \(b = 1\)

Substituting into the formula gives \((3y)^2 - 1^2\) which leads to the final product \(9y^2 - 1\). The method is simple, efficient, and works perfectly for special polynomial cases involving a difference of squares.
Understanding this pattern can make solving such expressions much faster and more intuitive.

Polynomials

Polynomials are expressions composed of variables and coefficients. They are summed together using mathematical operations such as addition, subtraction, multiplication, and non-negative integer exponents of variables.Let's break down the components:

• **Terms**: Parts of the polynomial separated by '+' or '-' signs.
• **Coefficients**: Numbers placed before variables (e.g., \(3\) in \(3y\)).
• **Degrees**: The highest exponent of the variable in the polynomial (e.g., the degree is \(2\) in \(9y^2\)).

In our case, we constructed another polynomial \(9y^2 - 1\) from multiplying two binomials. It's a polynomial because it includes sums and differences of variables raised to whole number exponents. It has two terms:

• \(9y^2\)
• \(1\)

By understanding the structure of polynomials, you can manipulate them in numerous ways, such as adding, subtracting, multiplying, and factoring polynomials. This makes it essential to grasp the parts and terminology regarding polynomials for algebra studies.

Algebraic Expressions

Algebraic expressions are like the building blocks of algebra. They consist of numbers, variables, and operations.Here’s how we define them:

• They can include operators like '+' and '-'.
• They often involve variables such as \(x\) or \(y\).
• They can be simple, like \(3x\) or more complex like \(9y^2 - 1\).

In the context of our example \((3y - 1)(3y + 1)\), each part is an algebraic expression itself. When we manipulate these expressions following algebraic rules, we derive meaningful relationships and solutions. Understanding algebraic expressions is important not just for simplifying them, but also for solving equations and tackling more advanced topics in mathematics. Remember that every term and every symbol has a role, and understanding that role is key to mastering algebra.