Problem 38 - Precalculus Mathematics for Calculus

Question

Multiply the algebraic expressions using a Special Product Formula and simplify. $$(2 y+5)(2 y-5)$$

Step-by-Step Solution

Answer

The simplified form is \(4y^2 - 25\).

Step 1: Identify the Special Product Formula

The given expression \((2y + 5)(2y - 5)\) can be recognized as a difference of squares formula, which is \((a + b)(a - b) = a^2 - b^2\). In this expression, \(a = 2y\) and \(b = 5\) are identified.

Step 2: Apply the Difference of Squares Formula

Using the formula \((a + b)(a - b) = a^2 - b^2\), substitute \(a = 2y\) and \(b = 5\) into the formula. This gives us: \((2y)^2 - 5^2\).

Step 3: Calculate Each Square Term

Calculate \((2y)^2\) and \(5^2\). We have \((2y)^2 = 4y^2\) and \(5^2 = 25\). So, the expression becomes \(4y^2 - 25\).

Step 4: Simplify the Expression

The expression \(4y^2 - 25\) is already simplified as there are no like terms to combine. Hence, \(4y^2 - 25\) is the simplified form of the original expression.

Key Concepts

Difference of Squares

The difference of squares is a specific type of algebraic identity that involves expressions of the form \((a + b)(a - b) = a^2 - b^2\). This formula is tremendously valuable for simplifying polynomial expressions, especially when dealing with binomials. The beauty of this formula lies in its simplicity and what it reveals about the symmetry of binomials, where the middle terms cancel each other out.
This makes it much easier to see the final result directly by just squaring the first term, squaring the second term, and then subtracting the latter from the former. In our given example, rather than expanding \((2y + 5)(2y - 5)\) into four individual multiplication steps, we use this shortcut to directly arrive at the simplified result, \(4y^2 - 25\).

• The square of \(a\) is \((2y)^2 = 4y^2\).
• The square of \(b\) is \(5^2 = 25\).
• Subtract the squares: \(4y^2 - 25\).

This formula saves time and reduces the potential for error during calculations.

Algebraic Expressions

Algebraic expressions form the backbone of algebra. They comprise numbers, variables, and operations (+, −, ×, ÷), combined into meaningful mathematical statements. Understanding them is key to solving various types of equations and inequalities.
In our context, an expression such as \((2y + 5)(2y - 5)\) is a classic example, involving a binomial multiplied by another binomial. With these, you can represent a multitude of situations and relationships in the world around us.

• Each part, like \(2y\) or \(5\), can vary depending on the value of the variable \(y\).
• Expressions allow for generalization in mathematics because they use variables, not fixed numbers.

With skills in manipulating algebraic expressions, students are empowered to solve more intricate problems, simply by recognizing patterns and rules like the special product formulas.

Polynomial Multiplication

Multiplying polynomials may initially appear complex, but it becomes quite manageable with practice and familiarity with certain formulas. One powerful tool is the application of special product formulas like the difference of squares. This method is particularly handy because it eliminates some of the cumbersome intermediary steps often involved in polynomial multiplication.
For our example, polynomial multiplication between \((2y + 5)\) and \((2y - 5)\) would traditionally involve using the distributive property:

• First, multiply \(2y\) by each term in the second binomial, \((2y - 5)\).
• Then, multiply \(5\) by each term in the second binomial, \((2y - 5)\).

Instead, by recognizing the expression as a difference of squares, we directly obtain \(4y^2 - 25\), streamlining the process. Mastering polynomial multiplication, especially through recognizing these patterns, enables students to handle larger and more complicated algebraic expressions efficiently.