Problem 66 - Elementary Algebra

Question

For the following problems, find the products. Expand \((a+b)(a-b)\) to prove it is equal to \(a^{2}-b^{2}\).

Step-by-Step Solution

Answer

Question: Expand the expression (a+b)(a-b) and prove that it is equal to a²-b².

Step 1: Distributive Property of Multiplication

Apply the distributive property of multiplication over addition to the given expression by multiplying each term within the first parentheses, \((a+b)\), with each term within the second parentheses, \((a-b)\).

Step 2: Multiply the First Terms

Multiply the first term of the first parentheses \((a)\) with the first term of the second parentheses \((a)\): \(a \times a = a^2\).

Step 3: Multiply the First and Second Terms

Multiply the first term of the first parentheses \((a)\) with the second term of the second parentheses \((-b)\): \(a \times (-b) = -ab\).

Step 4: Multiply the Second and First Terms

Multiply the second term of the first parentheses \((b)\) with the first term of the second parentheses \((a)\): \(b \times a = ab\).

Step 5: Multiply the Second Terms

Multiply the second term of the first parentheses \((b)\) with the second term of the second parentheses \((-b)\): \(b \times (-b) = -b^2\).

Step 6: Combine the Results

Combine the results from Steps 2 to 5: \(a^2- ab + ab - b^2\)

Step 7: Simplify the Expression

Observe that \(-ab\) and \(+ab\) cancel each other out, so the expression simplifies to: \(a^2 - b^2\) Thus, we have proved that the expansion of \((a+b)(a-b)\) is indeed equal to \(a^2 - b^2\).

Key Concepts

Distributive Property

Understanding the distributive property is critical when it comes to expanding algebraic expressions. This property tells us that if we have a term outside the parentheses, we can distribute it, or multiply it, to each term inside the parentheses. For example, when we have the expression \(a(b+c)\), we can apply the distributive property to expand it into \(ab + ac\).

In the exercise of proving the difference of squares, we apply this property twice. Once to distribute \(a\) over \(b+c\) and another time to distribute \(b\) over the same. It's like ensuring each friend gets an equal share of a pizza slice; every term inside the parentheses gets 'visited' by the term outside.

Algebraic Expansion

Algebraic expansion is essentially about 'stretching out' expressions, akin to unwinding a coiled spring. It involves the distributive property but also requires combining like terms to simplify the expression. When given the problem \( (a+b)(a-b) \), our goal through expansion is to eliminate the parentheses.

By multiplying each term from the first parenthesis with each term from the second parenthesis (as in our exercise), we effectively expand the expression. It's just like mixing ingredients separately before combining them to bake a cake—a step-by-step approach ensures everything is included.

Elementary Algebra

Elementary algebra provides the foundational building blocks we use when working with equations, expressions, and functions. It's the 'grammar' of the mathematical language. In our context, understanding how to manipulate variables and constants to simplify and solve equations is part of this core concept.

For instance, noticing that \( -ab \) and \( +ab \) cancel out in our solution is a direct application of elementary algebra. Developing this intuition is like learning to spot typos in a sentence—it makes the message clearer and the expression more concise.

Polynomial Multiplication

Polynomial multiplication is a way of 'growing' expressions where variables have exponents. It's foundational for advancing in algebra. When we multiply \( (a+b)(a-b) \) as in the provided exercise, we're doing more than multiplying numbers; we're combining terms with exponents.

The first and last terms \(a \times a\) and \(b \times -b\) give us \(a^2\) and \( -b^2\) respectively. It emphasizes that when we multiply terms with the same base, we add the exponents. Picture it as stacking blocks with numbers; when bases align, the heights—our exponents—add up.