Problem 45 - Algebra and Trigonometry Real Mathematics, Real People

Question

Use the Binomial Theorem to expand and simplify the expression. \(\left(\frac{1}{x}+y\right)^{5}\)

Step-by-Step Solution

Answer

The expanded and simplified form of the given expression using Binomial Theorem is \(32 + 160y/x + 240y^2/x^2 + 120y^3/x^3 + 20y^4/x^4 + y^5/x^5 \).

Step 1: Recognize the terms of the binomial

First identify the terms in the binomial. In our case, those are \( \frac{1}{x} \) and \( y \). The power to which the binomial is raised is \( 5 \). We'd be using these terms and the power to expand using the Binomial Theorem.

Step 2: Apply the Binomial Theorem

The expression can be expanded using the Binomial Theorem, which says \((a+b)^n = \sum_{k=0}^{n} {n \choose k}a^{n-k}b^k\). Substituting \(a=\frac{1}{x}\) and \(b=y\) into this theorem, the expression \(\left(\frac{1}{x}+y\right)^{5}\) can be expanded to: \[{5 \choose 0}\left(\frac{1}{x}\right)^5y^0 + {5 \choose 1}\left(\frac{1}{x}\right)^4y^1 + {5 \choose 2}\left(\frac{1}{x}\right)^3y^2 + {5 \choose 3}\left(\frac{1}{x}\right)^2y^3 + {5 \choose 4}\left(\frac{1}{x}\right)y^4 + {5 \choose 5}\left(\frac{1}{x}\right)^0y^5\]

Step 3: Calculate binomial coefficients

The binomial coefficients \({5 \choose k}\) can be calculated as:\( \frac{5!}{k!(5-k)!} \). Using these formulas to calculate the coefficients and substituting gives us the expansion: \[32x^{-5} + 160x^{-4}y + 240x^{-3}y^2 + 120x^{-2}y^3 + 20x^{-1}y^4 + y^5\]

Step 4: Simplify

The resulting polynomial expression can be simplified by factoring out \(x^5\) from every term. And thus, the simplified form of the binomial expansion is: \(32 + 160y/x + 240y^2/x^2 + 120y^3/x^3 + 20y^4/x^4 + y^5/x^5 \).

Key Concepts

Polynomial Expansion

Polynomial expansion involves expressing a power of a binomial, like \( \left(\frac{1}{x}+y\right)^{5} \), into a sum of terms. Each term is a product of powers of the individual parts of the binomial.

To expand a binomial, we use the Binomial Theorem.

• This theorem helps by providing a formula to express the power of a binomial as a series of terms.
• For a binomial \((a+b)^n\), the theorem states: \[ (a+b)^n = \sum_{k=0}^{n} {n \choose k} a^{n-k} b^k \] This means it is the sum of terms with specific coefficients.

In this exercise, \(a = \frac{1}{x}\) and \(b = y\). After expanding using this theorem,

you end up with a polynomial where each term is structured by multiplying a coefficient by the powers of the binomial’s components.

Binomial Coefficients

Binomial coefficients are critical in determining the weights of each term in the polynomial expansion. They are represented in the form \({n \choose k}\) and are calculated using the formula:\[ \frac{n!}{k!(n-k)!} \]

• The symbol \(!\) represents a factorial, which is the product of all positive integers up to that number.
• For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).

In our exercise, we expand \(\left(\frac{1}{x}+y\right)^{5}\),

so we compute coefficients for each term using this formula with \(n = 5\).
The coefficients for this expansion are \({5 \choose 0}, {5 \choose 1}, {5 \choose 2}, {5 \choose 3}, {5 \choose 4}, {5 \choose 5}\), resulting in the values \(1, 5, 10, 10, 5, 1\).

These tell us how much each term in our expanded polynomial is weighted.

Algebraic Expressions

Algebraic expressions involve numbers and variables combined using operations like addition, subtraction, multiplication, and division. In this exercise,

we manage the expression \(\left(\frac{1}{x}+y\right)^{5}\),

breaking it into simpler terms we can handle more easily.

• Each term evolves from the binomial expansion formula and involves powers of \(\frac{1}{x}\) and \(y\).
• This leads to terms like \(32\), \(160y/x\), \(240y^2/x^2\) etc., which are algebraic expressions themselves.

Simplifying these terms often involves combining like terms or factoring,

as in this case where \(32x^{-5}+160x^{-4}y\) is simplified to \(32 + 160\frac{y}{x}\).

This makes it easier to interpret and further use the expanded polynomial efficiently in any further algebraic work or calculus.