Problem 34 - Precalculus Mathematics for Calculus

Question

Multiply the algebraic expressions using a Special Product Formula and simplify. $$(r-2 s)^{2}$$

Step-by-Step Solution

Answer

The expression \((r-2s)^2\) simplifies to \(r^2 - 4rs + 4s^2\).

Step 1: Identify the Special Product Formula

The expression \((r-2s)^2\) represents a square of a binomial. We can apply the special product formula for the square of a binomial, which is \((a-b)^2 = a^2 - 2ab + b^2\). In this case, \(a = r\) and \(b = 2s\).

Step 2: Apply the Formula

Substitute \(a = r\) and \(b = 2s\) into the formula. So, \((r-2s)^2\) becomes \(r^2 - 2\cdot r \cdot 2s + (2s)^2\).

Step 3: Calculate Each Term

First, calculate \(r^2\), which remains \(r^2\). For the middle term, calculate \(-2 \cdot r \cdot 2s = -4rs\). Finally, calculate \((2s)^2 = 4s^2\).

Step 4: Combine the Terms

Add all the calculated terms together: \(r^2 - 4rs + 4s^2\). This is the simplified form of \((r-2s)^2\).

Key Concepts

Square of a Binomial

The concept of the "square of a binomial" is fundamental in algebra. A binomial is simply an expression that contains two terms. When we "square" a binomial, we are multiplying it by itself. This can often seem daunting, but using the special product formula simplifies the process.
For example, when you see

• extb{(a + b)^2}

you should automatically think of the formula: \[(a + b)^2 = a^2 + 2ab + b^2.\]Notice how the formula takes the binomial and expands it using three components:

• First, the square of the first term, \(a^2\)
• Second, two times the product of the two terms, \(2ab\)
• Finally, the square of the second term, \(b^2\)

By understanding this pattern, one can quickly find the square of any binomial.

Algebraic Expressions

Algebraic expressions involve numbers, variables, and arithmetic operations. They form the basis for nearly every problem in algebra. Understanding how to manipulate these expressions is key to mastering algebra.
An expression like \(r - 2s\)consists of two parts:

• \(r\) - a variable representing a number
• \(-2s\) - another term where \(-2\) is the coefficient, and \(s\) is the variable

These components combine with operations such as addition, subtraction, multiplication, and division. Remember, there's no "equals" sign in expressions, that would make it an equation!
Knowing how to write, interpret, and work with these expressions helps in breaking down complex problems into manageable parts. This is essential in both evaluating and simplifying expressions, allowing them to be easily understood and used for further calculations.

Simplifying Algebraic Expressions

Simplifying algebraic expressions means breaking them down into their simplest form. Often, this involves applying algebraic rules and formulas to combine like terms or use identities to reduce complexity.
For example, given the expression\((r-2s)^2\)One can use the formula for the square of a binomial to simplify it. Following the special product identity gives:\[r^2 - 4rs + 4s^2\]Observe how each term was carefully computed and then combined to arrive at a simpler expression. This process is called simplification, aiming to make an expression easier to work with or understand.
Always watch out for like terms—in this case, none can be simplified further since they are all distinct terms involving different combinations of\( r \) and \( s \). Successfully simplifying expressions means they are primed for further operations or analysis, showing their value in mathematics.