Problem 35 - Precalculus with Limits

Question

In Exercises 19 - 40, use the Binomial Theorem to expand and simplify the expression. \( \left(\dfrac{1}{x} + y\right)^5 \)

Step-by-Step Solution

Answer

The solution to \( \left(1/x + y\right)^5 \) is \( 1/x^5 + 5/x^4*y + 10/x^3*y^2 + 10/x^2*y^3 + 5/x*y^4 + y^5 \)

Step 1: Introduction to the Binomial Theorem

The Binomial Theorem is generally written as \[ (a + b)^n = \sum_{k=0}^{n} {n\choose k} a^{n-k} b^k \], where \({n\choose k}\) is the combination formula representing the number of ways to choose \(k\) items from \(n\) items. It shows how the terms of the expanded binomial will look: each term will have one component \(a^{n-k}\) that descends in degree from 'n' to '0' through the expansion, and one component \(b^k\) that ascends in degree from '0' to 'n'.

Step 2: Identifying 'a', 'b' and 'n'

From the given expression \(\left(1/x + y\right)^5\) , we can identify 'a' as \(1/x\), 'b' as \(y\), and 'n' as 5.

Step 3: Application of the Binomial Theorem

Applying the binomial theorem, we get \[ (a + b)^5 = {5\choose0} (1/x)^5 y^0 + {5\choose1} (1/x)^4 y^1 + {5\choose2} (1/x)^3 y^2 + {5\choose3} (1/x)^2 y^3 + {5\choose4} (1/x)^1 y^4 + {5\choose5} (1/x)^0 y^5 \]

Step 4: Computing combination values and simplifying

Computing the combination values, we get the following expanded form \[ = 1 * (1/x)^5 * y^0 + 5 * (1/x)^4 * y^1 + 10 * (1/x)^3 * y^2 + 10 * (1/x)^2 * y^3 + 5 * (1/x)^1 * y^4 + 1 * (1/x)^0 * y^5 \] Which simplifies to \[ = 1/x^5 + 5/x^4*y + 10/x^3*y^2 + 10/x^2*y^3 + 5/x*y^4 + y^5 \]

Key Concepts

Binomial Expansion

Understanding binomial expansion is crucial for simplifying expressions in algebra. It involves expanding a binomial expression, which is in the form of \(a + b)^n\), into a series of terms involving powers of \(a\) and \(b\) and specific coefficients.

The Binomial Theorem states that \( (a + b)^n = \sum_{k=0}^{n} {n\choose k} a^{n-k} b^k \), where the coefficient \( {n\choose k} \) is the number of combinations, representing how many ways we can select \(k\) items from a set of \(n\) without regard to order. These coefficients are also known as binomial coefficients and can be calculated using factorials or Pascal's Triangle.

A closer look at the theorem shows that each term in the binomial expansion is a combination of one part from 'a,' which descends in power, and one from 'b,' which ascends in power from one term to the next. By focusing on understanding and identifying these patterns, students can more easily expand binomials of any degree.

Combinations in Algebra

Combinations play a key role when working with binomial expansions. A combination, often denoted as \( {n\choose k} \), represents the number of ways to choose \(k\) elements from a set of \(n\) distinct elements without considering the order of selection.

For instance, when dealing with the binomial expansion of \( \left(\frac{1}{x} + y\right)^5 \), combinations tell us how many times each term will appear in the expansion. The formula for combinations is \( {n\choose k} = \frac{n!}{k!(n-k)!} \), where \( n! \) denotes the factorial of \( n \) and is a product of all positive integers less than or equal to \( n \).

Understanding how to compute these values is essential for working out the binomial coefficients and therefore correctly expanding and simplifying the polynomial. While it might seem complex, mastering the use of combinations in binomial expansion is a matter of practice and recognizing patterns.

Polynomial Simplification

Polynomial simplification is the process of reducing an expression to its simplest form. In the context of binomial expansion, this involves combining like terms and simplifying the coefficients.

After applying the Binomial Theorem to expand the binomial, the resulting expression contains various terms that may include fractions, powers, and products involving \( x \) and \( y \). To simplify these expressions, one must perform arithmetic operations: add, subtract, multiply, and divide, according to algebraic rules.

For example, in the expression \( 1/x^5 + 5/x^4\cdot y + 10/x^3\cdot y^2 + 10/x^2\cdot y^3 + 5/x\cdot y^4 + y^5 \) which was obtained from the binomial expansion of \( \left(\frac{1}{x} + y\right)^5 \), terms are already simplified, and there are no like terms to combine. In more complicated expansions, you might find terms that can be combined or perhaps coefficients that can be factored further, leading to a more compact and elegant expression.