Problem 23 - Algebra 2

Question

Expand each binomial. $$ (x+y)^{7} $$

Step-by-Step Solution

Answer

\(x^7 + 7x^6y + 21x^5y^2 + 35x^4y^3 + 35x^3y^4 + 21x^2y^5 + 7xy^6 + y^7\)

Step 1: Identify the Pattern

The expansion of \( (x+y)^n \) into a sum involving terms of the form \( ax^by^c \) is given by the formula: \( (x + y )^n = \sum_{k=0}^{n} {n \choose k} x^{n-k}y^k \), where {n \choose k} are the binomial coefficients. Here \( n \) is our power, 7.

Step 2: Perform the Expansion

Let's determine each term of the expansion following the given formula: \n Term 1: For k = 0, {7 \choose 0} x^{7-0}y^0 = 1*x^7*y^0 = x^7 \n Term 2: For k = 1, {7 \choose 1} x^{7-1}y^1 = 7*x^6*y = 7x^6y \n Term 3: For k = 2, {7 \choose 2} x^{7-2}y^2 = 21*x^5*y^2 = 21x^5y^2 \n Term 4: For k = 3, {7 \choose 3} x^{7-3}y^3 = 35*x^4*y^3 = 35x^4y^3 \n Term 5: For k = 4, {7 \choose 4} x^{7-4}y^4 = 35*x^3*y^4 = 35x^3y^4 \n Term 6: For k = 5, {7 \choose 5} x^{7-5}y^5 = 21*x^2*y^5 = 21x^2y^5 \n Term 7: For k = 6, {7 \choose 6} x^{7-6}y^6 = 7*x*y^6 = 7xy^6 \n Term 8: For k = 7, {7 \choose 7} x^{7-7}y^7 = 1*y^7 = y^7

Step 3: Write the Final Answer

Combining all the terms together, we get the final expanded binomial: \(x^7 + 7x^6y + 21x^5y^2 + 35x^4y^3 + 35x^3y^4 + 21x^2y^5 + 7xy^6 + y^7\)

Key Concepts

Binomial Coefficients

When expanding a binomial raised to a power, such as \((x + y)^7\), binomial coefficients come into play. These coefficients are the numbers in front of the terms in the expansion. They are derived from the combination formula \({n \choose k}\).
Binomial coefficients can be found using the formula:

• \({n \choose k} = \frac{n!}{k!(n-k)!}\)

where \(n!\) (n factorial) is the product of all positive integers up to \(n\), and \(k!\) is the product of all positive integers up to \(k\). This formula tells us how many ways we can choose \(k\) elements from a set of \(n\) elements,; thus, it's deeply rooted in combinatorics.
For example, if you're expanding \((x + y)^7\), the binomial coefficient for the first term is \(\binom{7}{0} = 1\), and their values move in sequence up to \(\binom{7}{7} = 1\). In this pattern, each coefficient gives the number of ways to arrange the terms of the binomial, multiplied by the corresponding powers of \(x\) and \(y\) in the expansion.

Polynomial Expansion

Polynomial expansion involves expressing expressions like \((x + y)^7\) as a sum of terms. These terms increase in complexity based on the power of the polynomial.
In the binomial expansion, each term is formed by:

• A binomial coefficient
• A power of \(x\)
• A power of \(y\)

Going step by step, we see that the expression \((x+y)^7\) is expanded into terms based on the binomial theorem, namely:\[x^7 + 7x^6y + 21x^5y^2 + 35x^4y^3 + 35x^3y^4 + 21x^2y^5 + 7xy^6 + y^7\]Here, each term comes from the different combinations of \(x\) and \(y\), with the coefficient in front determined by the number of ways to arrive at that particular arrangement. Each succeeding term increases the power of \(y\) while decreasing the power of \(x\), maintaining a constant sum of powers overall, which equals the original polynomial's power, 7 in this case.

Combinatorics

Combinatorics is a key principle used in the binomial theorem. It relates to counting the different ways to choose items from a group. When it comes to the binomial expansion, combinatorics helps us find the coefficients for each term.
The notation \({n \choose k}\) represents a combination, where:

• \(n\) is the total number of items
• \(k\) is the number of items to choose

The result gives the number of ways to select \(k\) items from \(n\). In our context, it's used to determine each coefficient of the terms in the polynomial expansion.
Why do we use combinatorics here? When expanding a binomial expression like \((x+y)^7\), for example, each term's arrangement depends on selecting one of these letters, and combinatorics gives us an organized yet simple way to count those arrangements efficiently. Thus, it's a crucial part of solving binomial expressions with both algebraic and real-world applications.

Algebraic Expressions

Algebraic expressions like \((x + y)^7\) consist of variables raised to powers, usually accompanied by constants or coefficients. In the binomial theorem, these expressions transform when expanded into a polynomial sum.
An algebraic expression is any mathematical phrase that includes:

• Numbers (constants)
• Variables (like \(x\) and \(y\))
• Operators (such as +, −, ×, ÷)

During expansion, each term in the algebraic expression is modified according to the rules of exponents and multiplication.
The given example, \((x + y)^7\), starts as a simple binomial algebraic expression and, through expansion, becomes a detailed polynomial expression. Each resulting term from the expansion contains variables raised to decreasing/increasing powers, illustrating how algebraic foundations transform complex expressions into simpler, workable components.