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Chapter 11: Problem 6

Expand. \((x+y)^{8}\)

Short Answer

Expert verified
\( (x + y)^8 = x^8 + 8x^7y + 28x^6y^2 + 56x^5y^3 + 70x^4y^4 + 56x^3y^5 + 28x^2y^6 + 8xy^7 + y^8 \).

Step by step solution

01

Understand the Binomial Theorem

The Binomial Theorem states that \[ (a + b)^n = \sum_{k=0}^{n} {n \choose k} a^{n-k} b^k \]where \[ {n \choose k} = \frac{n!}{k!(n-k)!} \] is the binomial coefficient.
02

Substitute Given Terms

For \( (x + y)^8 \), we have \( a = x \), \( b = y \), and \( n = 8 \). Thus the expansion is \[ (x + y)^8 = \sum_{k=0}^{8} {8 \choose k} x^{8-k} y^k \].
03

Calculate Binomial Coefficients

Calculate each binomial coefficient \( {8 \choose k} \) for \( k = 0 \) to \( k = 8 \):\[ \begin{align*} {8 \choose 0} &= 1, {8 \choose 1} &= 8, {8 \choose 2} &= 28, {8 \choose 3} &= 56, \ {8 \choose 4} &= 70, {8 \choose 5} &= 56, {8 \choose 6} &= 28, {8 \choose 7} &= 8, {8 \choose 8} &= 1 \end{align*} \]
04

Form the Expansion

Combine the binomial coefficients with powers of \( x \) and \( y \).\[ (x + y)^8 = 1x^8y^0 + 8x^7y^1 + 28x^6y^2 + 56x^5y^3 + 70x^4y^4 + 56x^3y^5 + 28x^2y^6 + 8x^1y^7 + 1x^0y^8 \].
05

Simplify

Simplify each term by removing powers of 1 and identifying coefficients:\[ (x + y)^8 = x^8 + 8x^7y + 28x^6y^2 + 56x^5y^3 + 70x^4y^4 + 56x^3y^5 + 28x^2y^6 + 8xy^7 + y^8 \].

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Key Concepts

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

Binomial Expansion
Let's start with the binomial expansion. This is a way to expand expressions that are raised to a power, such as \( (x + y)^8 \). The Binomial Theorem provides a formula to do this. It states that \[ (a + b)^n = \sum_{k=0}^{n} {n \choose k} a^{n-k} b^k \] where \[ {n \choose k} = \frac{n!}{k!(n-k)!} \] is the binomial coefficient. The sum goes from \( k = 0 \) to \( k = n \). Using this, we can expand any binomial expression easily.


So, for \( (x + y)^8 \), substitute \( a = x \), \( b = y \), and \( n = 8 \). This makes the expansion: \[ (x + y)^8 = \sum_{k=0}^{8} {8 \choose k} x^{8-k} y^k \].
Binomial Coefficient
Next, we need to understand binomial coefficients. These are the \( {n \choose k} \) terms in the binomial theorem. The binomial coefficient \( {n \choose k} = \frac{n!}{k!(n-k)!} \) tells us how many combinations of \( k \) items can be selected from \( n \) items.


For our example \( (x + y)^8 \), the binomial coefficients we need are: \[ \begin{align*} {8 \choose 0} &= 1, \ {8 \choose 1} &= 8, \ {8 \choose 2} &= 28, \ {8 \choose 3} &= 56, \ {8 \choose 4} &= 70, \ {8 \choose 5} &= 56, \ {8 \choose 6} &= 28, \ {8 \choose 7} &= 8, \ {8 \choose 8} &= 1 \end{align*} \] These coefficients will then be combined with the powers of \( x \) and \( y \) in the binomial expansion.
Polynomial Expansion
Once we have the binomial coefficients, we combine them with the terms \( x^{8-k} y^k \) to form the full polynomial expansion. Each term will take the form of \( {8 \choose k} x^{8-k} y^k \).


Let's put these together for \( (x + y)^8 \):

\[ (x + y)^8 = 1x^8y^0 + 8x^7y^1 + 28x^6y^2 + 56x^5y^3 + 70x^4y^4 + 56x^3y^5 + 28x^2y^6 + 8x^1y^7 + 1x^0y^8 \]


Simplifying each term by removing the powers of 1, we get:
\[ (x + y)^8 = x^8 + 8x^7y + 28x^6y^2 + 56x^5y^3 + 70x^4y^4 + 56x^3y^5 + 28x^2y^6 + 8xy^7 + y^8 \]

And there you have it! This is the polynomial expansion of \( (x + y)^8 \). Now, you can use the same process for any similar problems.

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