Problem 23 - A Graphical Approach to College Algebra

Question

Write the binomial expansion for each expression. $$(x+y)^{6}$$

Step-by-Step Solution

Answer

\((x+y)^6 = x^6 + 6x^5y + 15x^4y^2 + 20x^3y^3 + 15x^2y^4 + 6xy^5 + y^6\).

Step 1: Understand the Binomial Theorem

The binomial theorem states that \((x+y)^n\) can be expanded as \( \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^{k} \). This formula will be used to find the expansion of \((x+y)^6\).

Step 2: Identify the Values

In this case, \(n = 6\). We need to calculate the terms for \(k = 0\) to \(k = 6\), i.e., seven terms in total.

Step 3: Calculate Each Term

For each term in the expansion, calculate \(\binom{6}{k} x^{6-k} y^{k}\). We'll perform this calculation iteratively from \(k=0\) to \(k=6\).

Step 4: Calculate the Binomial Coefficients

The necessary binomial coefficients are: \(\binom{6}{0} = 1\), \(\binom{6}{1} = 6\), \(\binom{6}{2} = 15\), \(\binom{6}{3} = 20\), \(\binom{6}{4} = 15\), \(\binom{6}{5} = 6\), and \(\binom{6}{6} = 1\).

Step 5: Compute Each Term from Coefficients

Combine these coefficients with the corresponding power of \(x\) and \(y\):- For \(k=0\): \(\binom{6}{0} x^{6-0} y^{0} = x^6\)- For \(k=1\): \(\binom{6}{1} x^{6-1} y^{1} = 6x^5y\)- For \(k=2\): \(\binom{6}{2} x^{6-2} y^{2} = 15x^4y^2\)- For \(k=3\): \(\binom{6}{3} x^{6-3} y^{3} = 20x^3y^3\)- For \(k=4\): \(\binom{6}{4} x^{6-4} y^{4} = 15x^2y^4\)- For \(k=5\): \(\binom{6}{5} x^{6-5} y^{5} = 6xy^5\)- For \(k=6\): \(\binom{6}{6} x^{6-6} y^{6} = y^6\)

Step 6: Write the Complete Expression

Combine all the terms to write the complete expansion of \((x+y)^6\).The expansion is: \[x^6 + 6x^5y + 15x^4y^2 + 20x^3y^3 + 15x^2y^4 + 6xy^5 + y^6\]

Key Concepts

Binomial Theorem

The binomial theorem is a powerful tool in algebra that allows us to expand expressions raised to any power. When we see \((x+y)^n\), we can write it as a series of terms. Each term in this expansion involves combinations of \(x\) and \(y\). The general formula is given by:

\[ \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^{k} \]

Here, \(n\) is the exponent, \(x\) and \(y\) are the variables, and \(\binom{n}{k}\) is the binomial coefficient. This coefficient can be calculated using:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]

This theorem is fundamental in expanding binomial expressions effectively without directly multiplying each term.

Combinatorics

Combinatorics comes into play when calculating the binomial coefficients. In \((x+y)^6\), there are several coefficients, such as \(\binom{6}{0}, \binom{6}{1}, \binom{6}{2}\), and so on. These coefficients represent the number of ways to choose \(k\) items from \(n\) items without regard to order. Here's a quick guide:

• \(\binom{6}{0} = 1\): Choosing 0 items from 6 is just one way.
• \(\binom{6}{1} = 6\): Choosing 1 item from 6 can be done in 6 ways.
• \(\binom{6}{2} = 15\): Choosing 2 items can be done in 15 ways.
• Continuing this, you find \(\binom{6}{3} = 20\), \(\binom{6}{4} = 15\), and so forth.

These numbers become the coefficients of the terms in the expansion. Understanding how to calculate these is crucial for polynomial expansions.

Polynomial Expansion

Expanding a binomial like \((x+y)^6\) involves creating a polynomial with several terms. Each term is a combination of powers of \(x\) and \(y\), multiplied by a binomial coefficient. The process looks like this:

• Start with the highest power of \(x\), decreasing as you move forward.
• The power of \(y\) starts at zero and increases.
• Combine these with the respective coefficients.

For instance, the first term \(x^6\) comes from the coefficient \(\binom{6}{0}\), while the second term \(6x^5y\) uses \(\binom{6}{1}\). The complete expansion leads to:

\[ x^6 + 6x^5y + 15x^4y^2 + 20x^3y^3 + 15x^2y^4 + 6xy^5 + y^6 \]

This systematic approach helps break down the complexity into manageable parts.