Problem 35 - Precalculus Mathematics for Calculus

Question

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

Step-by-Step Solution

Answer

The expression simplifies to \(x^2 - 25\).

Step 1: Identify the Special Product Formula

The expression \((x+5)(x-5)\) is in the form \((a+b)(a-b)\), which is a difference of squares. This type of expression simplifies using the formula: \((a+b)(a-b) = a^2 - b^2\).

Step 2: Apply the Formula

Using the formula \((a+b)(a-b) = a^2 - b^2\), we identify \(a = x\) and \(b = 5\). Substitute these values into the formula: \((x+5)(x-5) = x^2 - 5^2\).

Step 3: Simplify the Expression

Now, simplify \(x^2 - 5^2\) by calculating the square of 5. This gives \(x^2 - 25\).

Step 4: Write the Final Simplified Expression

The simplified form of the expression \((x+5)(x-5)\) is \(x^2 - 25\), which is the final answer.

Key Concepts

Difference of Squares

The difference of squares is a special product formula in algebra. It allows you to easily simplify expressions of the form \((a+b)(a-b)\). This formula is handy for multiplying binomials quickly without having to use the long multiplication method. In this exercise, the expression \((x+5)(x-5)\) fits perfectly into the difference of squares formula: \((a+b)(a-b) = a^2 - b^2\). The idea behind this formula is when you take one binomial added and the other subtracted, the middle terms cancel each other out, leaving you only with the square of each term. This is why it is called 'difference of squares'. So for \((x+5)(x-5)\), fielding the formula gives \(x^2 - 25\) as a result. You'll see many algebra problems become simpler by recognizing this pattern.

Algebraic Expressions

An algebraic expression is a combination of numbers, variables, and operation symbols. These expressions represent real-world values and relationships. For instance, in \((x+5)(x-5)\), each bracket is an algebraic expression consisting of a variable \(x\) and constants. Algebraic expressions don't usually have an equals sign and aren't typically solved but are simplified or evaluated for given values of the variables. The main goal when dealing with expressions like \((x+5)(x-5)\) is to simplify them using known formulas or properties such as the difference of squares. Recognizing specific forms in algebraic expressions is key to using the right simplification methods efficiently.

Simplification Techniques

Simplification techniques are various methods used to make algebraic expressions easier to work with. Here, we focus on using special product formulas like the difference of squares to simplify the expression \((x+5)(x-5)\). Keeping an eye out for patterns helps simplify expressions quickly.

• Identify the structure: Recognize that \((x+5)(x-5)\) is in the form of \((a+b)(a-b)\), which fits the difference of squares.
• Apply the formula: Instantly apply the formula \(a^2 - b^2\) without expanding completely.
• Calculate squares: Be sure to compute any constants, like \(5^2\), to fully simplify the expression.

Simplification not only makes expressions manageable but also prepares them for further algebraic manipulation, if needed. Understanding when and how to use these techniques effectively allows for solving more complex problems with ease.