Problem 20 - College Algebra with Corequisite Support

Question

For the following exercises, use the Binomial Theorem to expand each binomial. $$ \left(\frac{1}{x}+3 y\right)^{5} $$

Step-by-Step Solution

Answer

\[ \left(\frac{1}{x} + 3y\right)^{5} = \frac{1}{x^5} + \frac{15y}{x^4} + \frac{90y^2}{x^3} + \frac{270y^3}{x^2} + \frac{405y^4}{x} + 243y^5 \]

Step 1: Identify the Binomial Formula

The Binomial Theorem states that \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). Where \( \binom{n}{k} \) is the binomial coefficient. For our binomial \( \left(\frac{1}{x} + 3y\right)^5\), we identify \(a = \frac{1}{x}\), \(b = 3y\), and \(n = 5\).

Step 2: Calculate Binomial Coefficients

Calculate the binomial coefficients \( \binom{5}{k} \) for \( k = 0, 1, 2, 3, 4, 5 \). These coefficients are: \( \binom{5}{0} = 1 \), \( \binom{5}{1} = 5 \), \( \binom{5}{2} = 10 \), \( \binom{5}{3} = 10 \), \( \binom{5}{4} = 5 \), \( \binom{5}{5} = 1 \).

Step 3: Expand the Binomial

Use the formula \( (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \) to expand the expression: \[ \left(\frac{1}{x} + 3y\right)^{5} = \sum_{k=0}^{5} \binom{5}{k} \left(\frac{1}{x}\right)^{5-k} (3y)^k \]Calculate each term:- For \(k=0\): \( \binom{5}{0} \left(\frac{1}{x}\right)^{5} (3y)^0 = \frac{1}{x^5} \)- For \(k=1\): \( \binom{5}{1} \left(\frac{1}{x}\right)^{4} (3y)^1 = 5 \left(\frac{1}{x^4}\right) 3y = \frac{15y}{x^4} \)- For \(k=2\): \( \binom{5}{2} \left(\frac{1}{x}\right)^{3} (3y)^2 = 10 \left(\frac{1}{x^3}\right) (9y^2) = \frac{90y^2}{x^3} \)- For \(k=3\): \( \binom{5}{3} \left(\frac{1}{x}\right)^{2} (3y)^3 = 10 \left(\frac{1}{x^2}\right) (27y^3) = \frac{270y^3}{x^2} \)- For \(k=4\): \( \binom{5}{4} \left(\frac{1}{x}\right)^{1} (3y)^4 = 5 \left(\frac{1}{x}\right) (81y^4) = \frac{405y^4}{x} \)- For \(k=5\): \( \binom{5}{5} \left(\frac{1}{x}\right)^{0} (3y)^5 = 1 (243y^5) = 243y^5 \)

Step 4: Write the Expanded Form

Combine all the calculated terms:\[ \left(\frac{1}{x} + 3y\right)^{5} = \frac{1}{x^5} + \frac{15y}{x^4} + \frac{90y^2}{x^3} + \frac{270y^3}{x^2} + \frac{405y^4}{x} + 243y^5 \]

Key Concepts

Binomial Coefficient

The binomial coefficient is a fundamental part of the Binomial Theorem, which is used to expand expressions like \((a + b)^n\). These coefficients are represented as \(\binom{n}{k}\) and are sometimes referred to as "combinations," indicating the number of ways to choose \(k\) elements from a total of \(n\) elements without regard to order. Their key role in algebra is to help determine the weight of each term in a binomial expansion.

• The binomial coefficient \(\binom{n}{k}\) can be calculated using the formula: \(\frac{n!}{k!(n-k)!}\), where \(!\) denotes factorial, a product of all positive integers less than or equal to a number.
• For example, to find \(\binom{5}{2}\), you compute \(\frac{5!}{2!(5-2)!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times 3 \times 2 \times 1} = 10\).

In the given problem, we compute binomial coefficients for \(k = 0, 1, 2, 3, 4, 5\) to determine the influence each term will have within the expanded polynomial.

Polynomial Expansion

The polynomial expansion involves using the Binomial Theorem to express a binomial raised to a power as a series of terms. Each term in the expansion is composed of the initial elements \(a\) and \(b\), raised to complementary powers, multiplied by the respective binomial coefficient.

• The general form of the expansion is \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\).
• This expansion process helps in breaking down complex expressions into simpler, more manageable parts.

For the given expression \(\left(\frac{1}{x} + 3y\right)^5\), this method allows us to convert it into a series form: every new term is a multiplication of a binomial coefficient, a power of \(\frac{1}{x}\), and a power of \(3y\). By calculating each term separately, we efficiently manage what might otherwise seem a daunting task, allowing clear, step-by-step simplification to reach the final expansion.

Coefficient Calculation

Calculating the coefficients of each term is an important step that involves several elements: the binomial coefficient, and the powers of both terms in the binomial expression. For each value of \(k\) in the expansion, you apply the formula \(\binom{5}{k} \left(\frac{1}{x}\right)^{5-k} (3y)^k\).

• Each coefficient is a product of: the selected binomial coefficient \(\binom{5}{k}\), the power of \(\frac{1}{x}\), and the power of \(3y\).
• For example, for \(k = 2\), the term calculates as \(10 \times \left(\frac{1}{x}\right)^3 \times 9y^2 = \frac{90y^2}{x^3}\).

Thus, for each step in the polynomial expansion, recalculating each coefficient ensures a correct, final expanded form. This part of the process confirms and solidifies understanding of how each component dilates the binomial into a full polynomial expression.