Problem 21 - Algebra and Trigonometry

Question

\(21-24\) Use the Binomial Theorem to expand the expression. $$ (x+2 y)^{4} $$

Step-by-Step Solution

Answer

The expanded form is \(x^4 + 8x^3y + 24x^2y^2 + 32xy^3 + 16y^4\).

Step 1: Identify Binomial Theorem

The Binomial Theorem states that \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). Here, \(a = x\), \(b = 2y\), and \(n = 4\). We will expand \((x + 2y)^4\) using this theorem.

Step 2: Calculate Binomial Coefficients

The coefficients are given by \(\binom{4}{k}\) for \(k = 0, 1, 2, 3, 4\). These are calculated as follows: \(\binom{4}{0} = 1\), \(\binom{4}{1} = 4\), \(\binom{4}{2} = 6\), \(\binom{4}{3} = 4\), \(\binom{4}{4} = 1\).

Step 3: Expand Each Term

Using the binomial coefficients, expand each term of the expression:- For \(k=0\): \( \binom{4}{0} x^{4} (2y)^0 = x^4 \).- For \(k=1\): \( \binom{4}{1} x^{3} (2y)^1 = 4x^3 \cdot 2y = 8x^3y\).- For \(k=2\): \( \binom{4}{2} x^{2} (2y)^2 = 6x^2 \cdot 4y^2 = 24x^2y^2\).- For \(k=3\): \( \binom{4}{3} x^{1} (2y)^3 = 4x \cdot 8y^3 = 32xy^3\).- For \(k=4\): \( \binom{4}{4} x^{0} (2y)^4 = 1 \cdot 16y^4 = 16y^4\).

Step 4: Combine Terms

Write the expanded form by combining all the terms obtained:The expansion is \(x^4 + 8x^3y + 24x^2y^2 + 32xy^3 + 16y^4\).

Key Concepts

Algebra

Algebra is a branch of mathematics dealing with symbols and the rules for manipulating those symbols. In this context, it helps us with expressing general forms and relationships between numbers. When we look at expressions like \((x + 2y)^4\), algebra provides the framework for expanding and simplifying them.

The essence of algebra in the Binomial Theorem is to recognize patterns and use those patterns to simplify expressions. It's all about finding ways to handle expressions that involve variables and understanding how they interact.
- Algebra allows us to solve equations by maintaining equality.
- It helps us to manipulate expressions such as \((x + 2y)^4\) into simpler forms.

By mastering algebra, you can easily tackle problems involving expansions, such as those solved through the Binomial Theorem.

Binomial Coefficients

Binomial coefficients are at the heart of the Binomial Theorem, representing the weights of each term in the expansion. For a given power \(n\), the binomial coefficient \(\binom{n}{k}\) is calculated as:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
This formula gives us the number of ways to choose \(k\) elements from \(n\) elements without considering the order.

In our example of \((x + 2y)^4\), binomial coefficients are:

• \(\binom{4}{0} = 1\)
• \(\binom{4}{1} = 4\)
• \(\binom{4}{2} = 6\)
• \(\binom{4}{3} = 4\)
• \(\binom{4}{4} = 1\)

These coefficients determine how each term in the binomial expansion is weighted.

Binomial coefficients also relate to Pascal's Triangle, another efficient way to find these numbers. Understanding these coefficients helps in evaluating the terms involved in any polynomial expansion.

Polynomial Expansion

Polynomial expansion involves expressing a power of a binomial as a sum of terms involving powers of the individual terms. In the example \((x + 2y)^4\), polynomial expansion breaks this expression down into a series of summed terms of the form \(a^{n-k}b^k\).

Using the Binomial Theorem, each term of the expansion for \((x + 2y)^4\) is calculated by:

• Recognizing the pattern of powers in the expansion \((x + 2y)^4 = x^4 + 8x^3y + 24x^2y^2 + 32xy^3 + 16y^4\)
• Using each corresponding binomial coefficient to determine the weight of each term
• Calculating each power, such as \((2y)^2\) to get \(4y^2\)

The expansion results in terms that are multiplied by their respective binomial coefficients and power combinations of \(x\) and \(y\).

Polynomial expansion allows us to break down and understand complex algebraic expressions in manageable parts, making calculations and simplifications much more straightforward.