Problem 41 - College Algebra

Question

Find the term indicated in each expansion. \((x-1)^{9} ;\) fifth term

Step-by-Step Solution

Answer

The fifth term in the expansion of \((x-1)^9\) is \(126x^5\).

Step 1: Calculating the combination

First calculate the combination \(_9C_4\), which gives us the coefficient of the term. This is done using the formula for combinations, which is \(^nC_r = \frac{n!}{r!(n-r)!}\) where n is the size of the set and r is the subset size. To find \(_9C_4\), it's \(\frac{9!}{4!(9-4)!}\) = 126.

Step 2: Power of the first term

Now calculate the power of the first term a in the binomial. According to the term finding formula, 'a' will have a power of \(n - k\). Here, n is 9 and k is 4. So, the power of 'a' is \(9 - 4\) = 5.

Step 3: Power of the second term

Next, find the power of the second term b in the binomial. According to the term finding formula, 'b' will have the power of k. Here, k is 4, so the power of 'b' is 4.

Step 4: Assemble the term

The final step is to put these values together. Using the formula, the 5th term is: \(_9C_4 \cdot x^{9-4} \cdot (-1)^4\). Plugging in our found values, it becomes \(126 \cdot x^5 \cdot 1 = 126x^5\)

Key Concepts

Combination Formula

The combination formula is a crucial tool in combinatorics, used to determine the number of ways you can choose a subset of items from a larger set. It's denoted as \(^nC_r\) and is calculated using the formula:

• \(^nC_r = \frac{n!}{r!(n-r)!}\)

This formula involves factorials, where \(n!\) (read as "n factorial") means the product of all positive integers up to \(n\). This helps in arranging complex selections.
Let's break it down with an example from our problem: we need \(_9C_4\).

• Calculate \(9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\)
• Calculate \(4! = 4 \times 3 \times 2 \times 1\) and \(5! = 5 \times 4 \times 3 \times 2 \times 1\)

Next, divide the results according to the formula:

• \(\frac{9!}{4! \times 5!} = \frac{362880}{24 \times 120} = 126\)

So, there are 126 ways to choose 4 items from a set of 9, which gives us our binomial coefficient.

Binomial Theorem

The Binomial Theorem provides a method to expand expressions that are raised to a power, in the form of

• \((a + b)^n\)

The expansion of this expression consists of several terms, each with a coefficient, powers of \(a\) and \(b\), and conforms to a pattern. The general term in the expansion is given by:

• \(_nC_k \cdot a^{n-k} \cdot b^k\)

Here, \(n\) is the exponent, \(k\) is the term number - 1, \(_nC_k\) is the combination coefficient, and the powers of \(a\) and \(b\) decrease and increase by 1 respectively with each term. This helps us find a particular term without expanding the whole expression.
In our exercise, we have \((x - 1)^9\), and we need the fifth term. The relevant values are \(n = 9\) and \(k = 4\). So, the term is:

• \(_9C_4 \cdot x^{9-4} \cdot (-1)^4 = 126 \cdot x^5 \cdot 1 = 126x^5\)

Thus, the Binomial Theorem efficiently delivers the target term.

Polynomial Expression

A polynomial expression is a sum of terms consisting of variables raised to whole number exponents and multiplied by coefficients. These expressions can model a wide variety of real-world phenomena and are structured as \(ax^n + bx^{n-1} + \ldots + z\), where each term resembles a "monomial."
In the context of the Binomial Theorem, each binomial raised to a power generates a polynomial as its expansion. For example, \((x-1)^9\) becomes a polynomial with multiple terms, each formed by different coefficients and powers of \(x\). This is vital for simplifying expressions and solving equations.
Particularly, binomials like \((x-1)^9\) can be transformed into a polynomial composed of terms like \(ax^m\), showing how powerful and adaptable polynomials are in both mathematics and its applications. Understanding polynomials paves the way to grasp more complex algebraic concepts, significantly enhancing problem-solving skills in mathematics.