Problem 23 - College Algebra with Modeling and Visualization

Question

Use the binomial theorem to expand each expression. $$ (p-q)^{6} $$

Step-by-Step Solution

Answer

The expansion is \( p^6 - 6p^5q + 15p^4q^2 - 20p^3q^3 + 15p^2q^4 - 6pq^5 + q^6 \).

Step 1: Understand the Binomial Theorem

The binomial theorem states that for any integer \( n \), \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\), where \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \). In this case, we will use \( a = p\), \( b = -q \), and \( n = 6 \).

Step 2: Identify Terms for Expansion

We are expanding \( (p-q)^6 \), which can be rewritten using the binomial theorem: \((p + (-q))^6\). This indicates that each term in the expansion will take the form \(\binom{6}{k} p^{6-k} (-q)^k\).

Step 3: Calculate Each Term of the Expansion

Now, we calculate each term from \( k = 0 \) to \( k = 6 \):1. For \( k = 0 \): \( \binom{6}{0} p^{6-0} (-q)^0 = 1 \cdot p^6 \cdot 1 = p^6 \)2. For \( k = 1 \): \( \binom{6}{1} p^{6-1} (-q)^1 = 6 \cdot p^5 \cdot (-q) = -6p^5q \)3. For \( k = 2 \): \( \binom{6}{2} p^{6-2} (-q)^2 = 15 \cdot p^4 \cdot q^2 = 15p^4q^2 \)4. For \( k = 3 \): \( \binom{6}{3} p^{6-3} (-q)^3 = 20 \cdot p^3 \cdot (-q)^3 = -20p^3q^3 \)5. For \( k = 4 \): \( \binom{6}{4} p^{6-4} (-q)^4 = 15 \cdot p^2 \cdot q^4 = 15p^2q^4 \)6. For \( k = 5 \): \( \binom{6}{5} p^{6-5} (-q)^5 = 6 \cdot p^1 \cdot (-q)^5 = -6pq^5 \)7. For \( k = 6 \): \( \binom{6}{6} p^{6-6} (-q)^6 = 1 \cdot p^0 \cdot q^6 = q^6 \)

Step 4: Combine All Terms

Combine all the terms calculated in the previous step:\( (p-q)^6 = p^6 - 6p^5q + 15p^4q^2 - 20p^3q^3 + 15p^2q^4 - 6pq^5 + q^6 \).

Key Concepts

Binomial Coefficient

The binomial coefficient is an essential concept in understanding the binomial theorem. For any integers \( n \) and \( k \), the binomial coefficient \( \binom{n}{k} \) determines the number of ways to choose \( k \) elements from a set of \( n \) elements without regard to the order of selection.
To calculate \( \binom{n}{k} \), use the formula:

• \( \frac{n!}{k!(n-k)!} \)

Here, the exclamation mark (!) denotes a factorial, which means multiplying a series of descending positive integers. For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).
Binomial coefficients play a crucial role in the expansion of binomial expressions and understanding their distribution, as shown in the problem with \( (p-q)^6 \). This result leads to terms weighted by these coefficients, such as 1, 6, 15, 20, 15, 6, and 1 in our example.

Expansion

In the context of the binomial theorem, expansion refers to expressing a binomial raised to a power as the sum of several terms. The binomial theorem formula is crucial and states:

• \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\)

When you expand \( (p-q)^6 \), it can be rewritten for simplicity as \((p+(-q))^6\), allowing us to apply the theorem.
Each term of the expansion represents a combination of powers of \( p \) and \( -q \), weighted by the binomial coefficient. This leads to a problem solution with terms like \( p^6 \), \(-6p^5q\), and more. The overall expression \((p-q)^6\) expands out in an organized way to reflect the different ways we can distribute the powers, providing both a systematic approach and deeper insights into algebraic manipulation.

Combinatorics

Combinatorics is the area of mathematics that involves counting, arranging, and finding patterns in sets of elements. It provides the mathematical foundation behind binomial coefficients.
In the binomial expansion, the coefficients \( \binom{n}{k} \) match the number of ways to choose \( k \) items from a set of \( n \), which is a core concept in combinatorics. This makes binomial expansions a fascinating case of applying combinatorial logic to algebra.
Moreover, in expansions like \((p-q)^6\), each coefficient mirrors the number of ways the respective term can appear when combining powers of \( p \) and \(-q\). This highlights the interplay between algebra and combinatorics, offering a great example of how these mathematical principles are interlinked with each other and assisting in solving complex algebraic expressions.