Problem 92 - Beginning and Intermediate Algebra

Question

Find the following special products. $$(5 y-4)(5 y+4)$$

Step-by-Step Solution

Answer

The special product of \((5y-4)(5y+4)\) is \(25y^2 - 16\).

Step 1: Identify terms for the difference of squares formula

The given expression can be represented as \((a-b)(a+b)\), where \(a = 5y\) and \(b = 4\).

Step 2: Apply the difference of squares formula

Now that we have identified \(a\) and \(b\), we can apply the difference of squares formula to the given expression: \((a-b)(a+b) = a^2 - b^2\).

Step 3: Calculate the special product

Substitute the values of \(a\) and \(b\) into the formula: \((5y - 4)(5y + 4) = (5y)^2 - (4)^2\) Now, simplify the expression: \((5y - 4)(5y + 4) = 25y^2 - 16\). The special product of \((5y - 4)(5y + 4)\) is \(25y^2 - 16\).

Key Concepts

Special Products

Special products in algebra are shortcuts that make multiplying certain types of polynomial expressions simpler. They allow us to quickly multiply pairs like

• conjugate binomials,
• squares of binomials,
• and others.

A common special product is the **difference of squares**. This is especially useful when dealing with expressions like \((a-b)(a+b)\). Using the difference of squares formula, we can instantly find the product as \(a^2 - b^2\). This is much faster than multiplying each term individually.
The key is recognizing these patterns and remembering their formulas, which makes solving these types of problems almost instant.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and arithmetic operations. In algebra, operations include addition, subtraction, multiplication, and division. Understanding algebraic expressions is crucial because they form the basis of many mathematical equations and expressions. When we talk about expressions like \((5y - 4)(5y + 4)\), we are dealing with terms and coefficients:

• "5y" is a term with "5" as the coefficient and "y" as the variable.
• "4" is a constant term.

Recognizing the parts of these expressions helps in applying formulas like the difference of squares, allowing us to solve for products and simplify quickly. Algebraic expressions often stand for quantities without fixed values, making them flexible and powerful in solving problems.

Polynomial Multiplication

Polynomial multiplication involves multiplying two polynomial expressions together. There are different methods to multiply polynomials, but recognizing patterns like

• **binomials and their special products**
aids greatly in simplifying the process. The multiplication is achieved by applying the distributive property: we distribute each term in the first polynomial by each term in the second.In the case of our expression \((5y - 4)(5y + 4)\), recognizing the special product allows us to skip lengthy terms multiplication and directly use \(a^2 - b^2\).
This gives us the product in fewer steps, avoiding error-prone operations. With practice, identifying these shortcuts when multiplying polynomials becomes second nature.