Problem 23 - Algebra and Trigonometry

Question

Expand each expression using the Binomial Theorem. $$ \left(x^{2}+y^{2}\right)^{5} $$

Step-by-Step Solution

Answer

\( (x^2 + y^2)^5 = x^{10} + 5x^8 y^2 + 10x^6 y^4 + 10x^4 y^6 + 5x^2 y^8 + y^{10} \)

Step 1 Title - Identify Components

Recognize that the expression \((x^2 + y^2)^5\) is in the form \((a + b)^n\). Here, \(a = x^2\), \(b = y^2\), and \(n = 5\).

Step 2 Title - Write Binomial Theorem

Recall the Binomial Theorem: \[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]

Step 3 Title - Apply the Binomial Theorem

Substitute \(a = x^2\), \(b = y^2\), and \(n = 5\) into the Binomial Theorem: \[ (x^2 + y^2)^5 = \sum_{k=0}^{5} \binom{5}{k} (x^2)^{5-k} (y^2)^k \]

Step 4 Title - Simplify Each Term

Simplify the exponentiated terms: \((x^2)^{5-k}\) becomes \(x^{2(5-k)}\) and \((y^2)^k\) becomes \(y^{2k}\). Therefore, \[ (x^2 + y^2)^5 = \sum_{k=0}^{5} \binom{5}{k} x^{2(5-k)} y^{2k} \]

Step 5 Title - Expand the Sum

Write out the sum for \(k = 0\) to \(k = 5\): \[(x^2 + y^2)^5 = \binom{5}{0} x^{10} y^0 + \binom{5}{1} x^8 y^2 + \binom{5}{2} x^6 y^4 + \binom{5}{3} x^4 y^6 + \binom{5}{4} x^2 y^8 + \binom{5}{5} x^0 y^{10}\]

Step 6 Title - Calculate Binomial Coefficients

Compute the binomial coefficients: \[ \binom{5}{0} = 1, \binom{5}{1} = 5, \binom{5}{2} = 10, \binom{5}{3} = 10, \binom{5}{4} = 5, \binom{5}{5} = 1 \]

Step 7 Title - Write Final Expanded Expression

Substitute the binomial coefficients into the expanded expression: \[(x^2 + y^2)^5 = 1 \cdot x^{10} + 5 \cdot x^8 y^2 + 10 \cdot x^6 y^4 + 10 \cdot x^4 y^6 + 5 \cdot x^2 y^8 + 1 \cdot y^{10} \]

Key Concepts

binomial theorem

The binomial theorem is a key concept in algebra that helps us expand expressions of the form \((a + b)^n\). It provides a way to expand these expressions without manually multiplying the terms repeatedly. This theorem states: \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). The symbol \binom{n}{k}\ represents binomial coefficients, another vital part of the theorem.
Using the binomial theorem makes it easier to handle larger exponents and complex polynomial expressions. For example, in our exercise, we expand \( (x^2 + y^2)^5 \) by following this principle.
To utilize the theorem, recognize the components \(a\), \(b\), and \(n\) from the expression. Then use the summation formula to find each term in the expansion.

binomial coefficients

Binomial coefficients are the numerical factors in the binomial theorem expansion and are denoted as \binom{n}{k}\. They can be calculated using the formula: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] Here, the factorial symbol (!) means to multiply all positive integers up to that number.
For example, in the exercise, we need the coefficients when \( n = 5 \). We compute:

• \binom{5}{0} = 1\
• \binom{5}{1} = 5\
• \binom{5}{2} = 10\
• \binom{5}{3} = 10\
• \binom{5}{4} = 5\
• \binom{5}{5} = 1\

These coefficients help us form the individual terms of the expanded polynomial. For instance, multiplying \binom{5}{2} = 10\ by the variables’ respective powers in our formula gives us the term \10 x^6 y^4\.

polynomial expansion

Polynomial expansion using the binomial theorem involves breaking down a polynomial raised to a power into a sum of terms. Each term is a product of binomial coefficients and the variables raised to appropriate powers.
For the expression \( (x^2 + y^2)^5 \), applying the binomial theorem, we get:
\ (x^2 + y^2)^5 = \sum_{k=0}^{5} \binom{5}{k} (x^2)^{5-k} (y^2)^k \.
Simplified, this becomes: \ \binom{5}{0} x^{10} y^0 + \binom{5}{1} x^8 y^2 + \binom{5}{2} x^6 y^4 + \binom{5}{3} x^4 y^6 + \binom{5}{4} x^2 y^8 + \binom{5}{5} x^0 y^{10} \.
Finally, substituting the calculated binomial coefficients:
\ (x^2 + y^2)^5 = 1 \cdot x^{10} + 5 \cdot x^8 y^2 + 10 \cdot x^6 y^4 + 10 \cdot x^4 y^6 + 5 \cdot x^2 y^8 + 1 \cdot y^{10} \.
This expansion gives us a simplified polynomial that is much easier to handle analytically or computationally.