Problem 52 - College Algebra

Question

Find each product. $$(x+2)^{3}$$

Step-by-Step Solution

Answer

The cube of the binomial \( (x+2)^{3} \) is \(x^{3}+6x^{2}+12x+8\).

Step 1: Expand Using Repeated Multiplication

Start by writing out the cube as repeated multiplication. In other words, write \((x+2)^{3} \) as \((x+2)*(x+2)*(x+2) \)

Step 2: Multiply the First Two Binomials

Multiply the first two binomials to get the intermediate product. So \((x+2)*(x+2) \) becomes \(x^{2}+4x+4\) which simplifies the next step.

Step 3: Multiply the Last Binomial

Then you need to multiply out \(x^{2}+4x+4 \) and \(x+2) \), which will give \(x^{3}+6x^{2}+12x+8\)

Step 4: Write Down the Final Result

After simplification of the expressions from above steps, you will have \(x^{3}+6x^{2}+12x+8\)

Key Concepts

Polynomial Multiplication

Polynomial multiplication is a process that involves taking one polynomial and multiplying it with another. This is a crucial skill needed in algebra to simplify and solve various problems. Imagine you have two expressions, each consisting of multiple terms (or monomials), much like the expression in our example exercise:

• The expression \((x+2)^{3}\) essentially requires multiplying three identical binomials, \(x+2\).
• To make this simpler, it's broken down into stages: First, multiply two of the binomials; then, take the result and multiply it by the third.
• The associative property of multiplication is our friend here, as it lets us multiply in steps without changing the overall outcome.

When you multiply two polynomials, each term in the first polynomial multiplies every term in the second. You end up with a new polynomial with terms that you then combine and simplify by adding.

Multiplying the binomials involves breaking them into parts and applying the distributive law, which helps keep things organized as you work with larger expressions.

Algebraic Expressions

Algebraic expressions are a fundamental part of algebra and mathematics in general. They are combinations of numbers, variables, and operation signs that represent specific values or quantities. In this context, variables are symbols like \(x\), which could represent any number. An algebraic expression such as \(x+2\) is straightforward, consisting of the variable \(x\) and the constant \(2\).

In our given exercise:

• We have an algebraic expression raised to the power of three, indicating the puzzle is about expanding \(x+2\) three times.
• Expressions can take various forms, such as binomials, trinomials, and more complex polynomials, depending on the number of terms.

The goal when dealing with algebraic expressions, especially in exercises like our example, is to simplify them as much as possible. This simplification might involve operations like adding, subtracting, and distributing for easier manipulation and eventually obtaining a clean result.

Understanding how to handle these expressions and the rules that govern them is essential for solving many algebraic problems.

Cube of a Binomial

The cube of a binomial is a specific case of polynomial multiplication where you multiply a binomial by itself twice. For instance, consider the expression \( (x+2)^3 \). This means \( (x+2) \) is multiplied by itself twice more, making a total of three multiplications:

• Firstly, the first two \( (x+2) \) terms are multiplied together to get an intermediate result: \( x^2 + 4x + 4 \).
• This intermediate product is then multiplied by the third \( (x+2) \) term.
• The multiplication involves using distributive and associative properties, ensuring each term from the polynomials is correctly accounted for in the final output.

Understanding binomial expansion is useful because many algebraic tasks involve simplifying powers of binomials. The general formula for any binomial raised to any power, often called the Binomial Theorem, provides coefficients that tell us how to expand any \( (a+b)^n \). However, in this problem, we handle it manually to truly see how each term contributes to the final polynomial.

By mastering how to cube a binomial, you lessen the challenge of handling more complex polynomial manipulations.