Problem 72 - 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: Identify the Binomial Theorem

The binomial theorem states that \[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k. \] In this problem, we have \(a = x \), \(b = 2y \), and \(n = 5.\)

Step 2: Apply the Binomial Theorem

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

Step 3: Compute the Binomial Coefficients

Calculate the binomial coefficients \( \binom{5}{k} \) for \( k = 0, 1, 2, 3, 4, 5\): \[\begin{align*} \binom{5}{0} &= 1, \ \binom{5}{1} &= 5, \ \binom{5}{2} &= 10, \ \binom{5}{3} &= 10, \ \binom{5}{4} &= 5, \ \binom{5}{5} &= 1 \end{align*}\]

Step 4: Expand Each Term

Using the binomial coefficients, expand each term: \[(x + 2y)^5 = 1 \cdot x^5 \cdot (2y)^0 + 5 \cdot x^4 \cdot (2y)^1 + 10 \cdot x^3 \cdot (2y)^2 + 10 \cdot x^2 \cdot (2y)^3 + 5 \cdot x \cdot (2y)^4 + 1 \cdot x^0 \cdot (2y)^5\]

Step 5: Simplify Each Term

Simplify each expanded term: \[\begin{align*} x^5 \cdot 1 &= x^5, \ 5x^4 \cdot 2y &= 10x^4y, \ 10x^3 \cdot (2y)^2 &= 40x^3y^2, \ 10x^2 \cdot (2y)^3 &= 80x^2y^3, \ 5x \cdot (2y)^4 &= 80xy^4, \ (2y)^5 &= 32y^5 \end{align*}\]

Step 6: Combine the Simplified Terms

Combine all the simplified terms to get the final expanded form: \[(x + 2y)^5 = x^5 + 10x^4y + 40x^3y^2 + 80x^2y^3 + 80xy^4 + 32y^5.\]

Key Concepts

Algebra

Algebra forms the foundation of many areas in mathematics. When working with algebra, you manipulate symbols and letters to represent numbers and express relationships. These manipulations often involve operations such as addition, subtraction, multiplication, and division.
In the context of binomial expansions, algebra helps us understand how to distribute terms, simplify expressions, and solve for unknowns. The binomial theorem is an example of how algebraic principles are applied to expand expressions of the form \((a + b)^n\).

When breaking down complex problems like \((x + 2y)^5\), algebra helps us systematically apply the binomial theorem to achieve a simplified expanded form. Remember, mastering algebraic techniques like factoring, combining like terms, and applying formulas will be beneficial in both this context and broader mathematical explorations.

Binomial Coefficients

Binomial coefficients are key components in the expansion of binomials. They are represented as \[ \binom{n}{k} \] and read as 'n choose k'. These coefficients indicate the number of ways to choose k elements from a set of n elements without regard to the order.

The formula for binomial coefficients is:
\ \binom{n}{k} = \frac{n!}{k!(n-k)!} \ \.

In the given exercise of expanding \((x + 2y)^5\), we calculate the coefficients for each term as follows:
\ \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 are then used to multiply the corresponding terms in the expansion. Practicing the use of binomial coefficients can simplify many problems in combinatorics and probability.

Keep practicing how to find and use binomial coefficients effectively – it’s an essential skill in algebra and higher mathematics.

Expansion

The process of expansion in algebra, particularly with the binomial theorem, involves transforming expressions like \((a + b)^n\) into a sum of terms. Each term in the expansion corresponds to a specific combination of the binomial’s elements raised to respective powers and multiplied by the binomial coefficients.

Let’s break down the expansion of \((x + 2y)^5\):

• Identify the coefficients using binomial coefficients.
• Substitute the elements (x and 2y) and their corresponding powers into each term.
• Multiply the coefficients and simplify each term.

The step-by-step solution provided illustrates this process, expanding and simplifying to obtain:
\ x^5 + 10x^4y + 40x^3y^2 + 80x^2y^3 + 80xy^4 + 32y^5 \
Expansion is useful in many domains, including series calculation, probability, and algebraic proofs. Understanding the detailed process helps ensure accuracy and efficiency in solving similar problems.