Problem 16 - College Algebra with Corequisite Support

Question

For the following exercises, use the Binomial Theorem to expand each binomial. $$ (2 x+3 y)^{4} $$

Step-by-Step Solution

Answer

\(16x^4 + 96x^3y + 216x^2y^2 + 216xy^3 + 81y^4\)

Step 1: Understand the Binomial Theorem

The binomial theorem states that for any positive integer \( n \), \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). In this case, \( a = 2x \) and \( b = 3y \), and \( n = 4 \).

Step 2: Set up the Expression Using Binomial Coefficients

The expression \((2x + 3y)^4\) can be expanded by calculating each term in the sum: \[(2x + 3y)^4 = \sum_{k=0}^{4} \binom{4}{k} (2x)^{4-k} (3y)^k\].

Step 3: Compute Each Term of the Expansion

We need to calculate each term from \( k = 0 \) to \( k = 4 \). The general term is \( \binom{4}{k} (2x)^{4-k} (3y)^k \).

Step 4: Calculate Individual Terms

- When \( k = 0 \): \[\binom{4}{0} (2x)^4 (3y)^0 = 1 \cdot 16x^4 \cdot 1 = 16x^4\].- When \( k = 1 \): \[\binom{4}{1} (2x)^3 (3y)^1 = 4 \cdot 8x^3 \cdot 3y = 96x^3y\].- When \( k = 2 \): \[\binom{4}{2} (2x)^2 (3y)^2 = 6 \cdot 4x^2 \cdot 9y^2 = 216x^2y^2\].- When \( k = 3 \): \[\binom{4}{3} (2x)^1 (3y)^3 = 4 \cdot 2x \cdot 27y^3 = 216xy^3\].- When \( k = 4 \): \[\binom{4}{4} (2x)^0 (3y)^4 = 1 \cdot 1 \cdot 81y^4 = 81y^4\].

Step 5: Write the Final Expanded Expression

Combine all terms calculated in Step 4: \[16x^4 + 96x^3y + 216x^2y^2 + 216xy^3 + 81y^4\].

Key Concepts

Binomial Expansion

The process of binomial expansion allows us to expand expressions that are raised to a power. This plays a crucial role in algebra, helping us to deal with expressions like \((a + b)^n\)\.,where \(a\) and \(b\) are any numbers or variables, and \(n\) specifies the power.
To successfully expand such expressions, we rely on the Binomial Theorem. This theorem provides a structured way to predict the expansion by using binomial coefficients.
According to the theorem, \((a + b)^n\) can be expanded as:\[\sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\].In our specific example of \((2x + 3y)^4\),the theorem guides us to expand the expression into multiple terms that reflect the individual influence of each power combination of \(2x\) and \(3y\). By using coefficients and varying powers of each term, the expansion goes from having a squared term to having an expression fully broken down into its basic polynomial terms.

Algebraic Expressions

An algebraic expression is a collection of numbers, variables, and mathematical operations that form a relation or an equation.
During binomial expansion, our job is to take compound algebraic expressions and simplify them into a sequence of simpler terms.
In the context of our original exercise, \((2x + 3y)^4\),we must identify \(2x\) as one term and \(3y\) as another, both being raised together to the power of four.
This means we systematically adjust the powers of \(2x\) and \(3y\) in each term of the expansion, ensuring that each combination of their powers adds up to \(n\), which is 4 in this case. This adjustment freshly rearranges the expression into a more analytics-friendly form by breaking it down to base parts.
Understanding the structure of algebraic expressions being expanded is crucial, as it influences how we manipulate powers and apply calculations based on binomial coefficients.

Coefficient Calculation

In any binomial expansion, the coefficients play a pivotal role. They are the numbers in front of the algebraic terms and are derived from combinations in math.
The coefficients within our binomial expansion use binomial coefficients represented by \(\binom{n}{k}\). This symbol refers to the number of ways to choose \(k\) elements from a set of \(n\) elements without regard to order, often called "n choose k."
In calculations, these coefficients are paramount in distributing the terms correctly.
For example, in \((2x + 3y)^4\),when calculating individual terms:

• The coefficient for \(k = 0\) is \(\binom{4}{0} = 1\), leading to the term \(16x^4\).
• When \(k = 1\), the coefficient becomes \(\binom{4}{1} = 4\), resulting in the term \(96x^3y\).
• With \(k = 2\), we have \(\binom{4}{2} = 6\), yielding the term \(216x^2y^2\).
• For \(k = 3\), \(\binom{4}{3} = 4\) defines the term \(216xy^3\).
• Finally, \(k = 4\) gives us \(\binom{4}{4} = 1\), which computes to \(81y^4\).

Understanding how to calculate the coefficient correctly is critical to correctly executing binomial expansions and achieving accurate results in polynomial form.