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Chapter 6: Problem 5

Using the binomial theorem, expand each. $$(x+y)^{4}$$

Short Answer

Expert verified
Using the binomial theorem, the expansion of \((x+y)^{4}\) is: $$ (x+y)^{4} = x^{4} + 4x^{3}y + 6x^{2}y^{2} + 4xy^{3} + y^{4} $$

Step by step solution

01

Identify the value of n, a, and b

In our given expression \((x+y)^{4}\), we have \(n=4\), \(a=x\), and \(b=y\).
02

Calculate the binomial coefficients

We will now calculate the binomial coefficients \({4 \choose k}\) for each term in the sum: - For \(k=0\): \({4 \choose 0} = \frac{4!}{0!(4-0)!} = \frac{4!}{0!4!} = 1\) - For \(k=1\): \({4 \choose 1} = \frac{4!}{1!(4-1)!} = \frac{4!}{1!3!} = 4\) - For \(k=2\): \({4 \choose 2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = 6\) - For \(k=3\): \({4 \choose 3} = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = 4\) - For \(k=4\): \({4 \choose 4} = \frac{4!}{4!(4-4)!} = \frac{4!}{4!0!} = 1\)
03

Expand the sum using the binomial theorem

Now, we will use the binomial theorem to write out the expanded expression: \((x+y)^{4} = \sum_{k=0}^{4} {4 \choose k} x^{4-k} y^{k}\) Plug in the calculated binomial coefficients from step 2: \((x+y)^{4} = {4 \choose 0}x^{4}y^{0} + {4 \choose 1}x^{3}y^{1} + {4 \choose 2}x^{2}y^{2} + {4 \choose 3}x^{1}y^{3} + {4 \choose 4}x^{0}y^{4}\) Finally, substitute the calculated coefficients: \((x+y)^{4} = 1x^{4}y^{0} + 4x^{3}y^{1} + 6x^{2}y^{2} + 4x^{1}y^{3} + 1x^{0}y^{4}\) Simplify the powers of x and y: \((x+y)^{4} = x^{4} + 4x^{3}y + 6x^{2}y^{2} + 4xy^{3} + y^{4}\) So, the expanded expression of \((x+y)^{4}\) is: $$ (x+y)^{4} = x^{4} + 4x^{3}y + 6x^{2}y^{2} + 4xy^{3} + y^{4} $$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binomial Coefficients
Binomial coefficients are essential in the context of the binomial theorem, representing the number of ways to choose elements from a set. These coefficients are represented as \({n \choose k}\), which translates to the formula: \[{n \choose k} = \frac{n!}{k!(n-k)!}\]
The exclamation mark "!" represents a factorial, meaning the product of all positive integers up to that number. For instance, \(4! = 4 \times 3 \times 2 \times 1 = 24\).
In the binomial expansion of \((x+y)^4\), we compute coefficients like:
  • \({4 \choose 0} = 1\)
  • \({4 \choose 1} = 4\)
  • \({4 \choose 2} = 6\)
  • \({4 \choose 3} = 4\)
  • \({4 \choose 4} = 1\)
These coefficients are vital in constructing each term of the polynomial.
Polynomial Expansion
Polynomial expansion involves expressing a binomial raised to a power as a sum of terms. This is beautifully illustrated through the binomial theorem:\[(x+y)^n = \sum_{k=0}^{n} {n \choose k} x^{n-k} y^k\]
When expanding \((x+y)^4\), each term is calculated by using the binomial coefficients as multipliers:
  • The first term is \({4 \choose 0}x^4y^0 = x^4\)
  • The second term is \({4 \choose 1}x^3y^1 = 4x^3y\)
  • The third term is \({4 \choose 2}x^2y^2 = 6x^2y^2\)
  • The fourth term is \({4 \choose 3}x^1y^3 = 4xy^3\)
  • The fifth term is \({4 \choose 4}x^0y^4 = y^4\)
Combining these terms gives us the complete expanded form of the polynomial: \(x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4\).
Combinatorics
Combinatorics is the branch of mathematics concerned with counting, arrangement, and combination of elements. In relation to the binomial theorem, it helps understand the calculation of binomial coefficients.
The specifics of combinatorics involve:
  • Understanding the concept of "n choose k," which is a combination formula denoting how to choose \(k\) elements from a total of \(n\) without regard to the order.
  • Applying factorial notation in calculations, ensuring you know how to simplify expressions like \(\frac{4!}{2!2!} = 6\).
Combinatorics provides the tools to explore deeper problems involving permutations, combinations, and arrangements, vital for mastering the binomial theorem and its vast applications.

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