Problem 73 - Algebra and Trigonometry

Question

Based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Use the binomial theorem to expand: \((x+2 y)^{5}\)

Step-by-Step Solution

Answer

(x + 2y)^5 = x^5 + 10x^4y + 40x^3y^2 + 80x^2y^3 + 80xy^4 + 32y^5

Step 1 - Understand the Binomial Theorem

The binomial theorem provides a formula for expanding expressions of the form \((a + b)^n\). The formula is given by: \[ (a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^{k} \] Here, \(\binom{n}{k}\) is the binomial coefficient, \(a\) and \(b\) are the terms in the binomial, and \(n\) is the exponent.

Step 2 - Identify Parts of the Binomial

In the expression \((x + 2y)^5\), identify \(a\), \(b\), and \(n\) based on the binomial theorem formula. Here, \(a = x\), \(b = 2y\), and \(n = 5\).

Step 3 - Write Down the General Term

The general term for the expansion using the binomial theorem is: \[\binom{n}{k} a^{n-k} b^k \] Substitute \(a = x\), \(b = 2y\), and \(n = 5\): \[\binom{5}{k} x^{5-k} (2y)^k \]

Step 4 - Calculate Each Term

Calculate each term by substituting values of \(k\) from 0 to 5: \[ \begin{aligned} &\text{For } k = 0: \binom{5}{0} x^{5-0} (2y)^{0} = 1 \times x^5 \times 1 = x^5 \ &\text{For } k = 1: \binom{5}{1} x^{5-1} (2y)^{1} = 5 \times x^4 \times 2y = 10x^4y \ &\text{For } k = 2: \binom{5}{2} x^{5-2} (2y)^{2} = 10 \times x^3 \times 4y^2 = 40x^3y^2 \ &\text{For } k = 3: \binom{5}{3} x^{5-3} (2y)^{3} = 10 \times x^2 \times 8y^3 = 80x^2y^3 \ &\text{For } k = 4: \binom{5}{4} x^{5-4} (2y)^{4} = 5 \times x \times 16y^4 = 80xy^4 \ &\text{For } k = 5: \binom{5}{5} x^{5-5} (2y)^{5} = 1 \times 1 \times 32y^5 = 32y^5 \ \ \text{So, the expanded form is:} \ (x + 2y)^5 = x^5 + 10x^4y + 40x^3y^2 + 80x^2y^3 + 80xy^4 + 32y^5 \ \ \ \end{aligned} \]

Key Concepts

Binomial Expansion

The concept of binomial expansion allows us to expand expressions that are raised to a power. Using the binomial theorem, we can break down expressions like \(a + b\)^n into simpler components. This is particularly useful in algebra when dealing with higher powers of polynomials.

Binomial Coefficient

The binomial coefficient, represented by \binom{n}{k}\, plays a crucial role in binomial expansion. It indicates the number of ways to choose k elements from a set of n elements. The coefficient itself can be computed using the formula: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] where \!\ denotes factorial, the product of all positive integers up to that number.

Polynomial Expansion

Polynomial expansion involves expressing a binomial raised to a power as a sum of individual terms. Each term consists of a binomial coefficient and the product of the two variables raised to their respective powers. For example, in the expression \(x + 2y\)^5, we expand it to get: \x^5 + 10x^4y + 40x^3y^2 + 80x^2y^3 + 80xy^4 + 32y^5.

Algebra

Algebra often requires the application of binomial and polynomial expansions to simplify and solve equations. Understanding these concepts allows students to handle complex algebraic expressions efficiently. By breaking down terms, finding coefficients, and simplifying results, algebra becomes more manageable and easier to understand.