Problem 50 - College Algebra with Modeling and Visualization

Question

Find the specified term. The eighth term of \((2 a-b)^{9}\)

Step-by-Step Solution

Answer

The eighth term is \(-144a^2b^7\).

Step 1: Understanding the Problem

We are tasked with finding the eighth term in the expansion of \((2a - b)^9\). This is a binomial expression, so we can use the Binomial Theorem to find a specific term.

Step 2: Binomial Theorem Formula

The Binomial Theorem tells us that any term in the expansion of \((x+y)^n\) is given by \[ T_{k} = \binom{n}{k} \cdot x^{n-k} \cdot y^k \] where \( \binom{n}{k} \) is the binomial coefficient, \( x \) and \( y \) are terms in the binomial, and \(n\) is the exponent.

Step 3: Identify Parameters

Here, \( n = 9 \), \( x = 2a \), and \( y = -b \). We need to find the eighth term which corresponds to \( k = 7 \) (since term counting starts with 0).

Step 4: Calculating the Binomial Coefficient

The binomial coefficient \( \binom{9}{7} \) is calculated as follows:\[ \binom{9}{7} = \binom{9}{2} = \frac{9 \times 8}{2 \times 1} = 36 \]

Step 5: Calculate the Term

Substituting \( x = 2a \), \( y = -b \), \( n = 9 \), and \( k = 7 \) into the formula:\[ T_7 = \binom{9}{7} \times (2a)^{9-7} \times (-b)^7 \] This simplifies to:\[ T_7 = 36 \times (2a)^2 \times (-b)^7 \] Calculating each part gives:\[ 36 \times 4a^2 \times -b^7 = 36 \times 4 \times a^2 \times (-b)^7 = -144a^2b^7 \]

Step 6: Conclusion

Therefore, the eighth term of the expansion is \(-144a^2b^7\).

Key Concepts

Binomial Expansion

The Binomial Expansion is a method used in algebra to express the power of a binomial, that is an expression consisting of two terms, raised to a positive integer power. It provides a way to expand expressions like \( (x + y)^n \) into a series of terms. Each term in this expansion is a combination of the powers of \( x \) and \( y \).

To employ the Binomial Expansion, you rely on the Binomial Theorem, which gives you a formula to find any term within the expansion without having to multiply the binomial out completely. This makes it incredibly useful when dealing with high powers. By simplifying the process of finding individual terms, the Binomial Expansion opens up doors for more efficient calculations in polynomial algebra.

Binomial Coefficient

In the context of the Binomial Expansion, the Binomial Coefficient plays a crucial role. It is denoted by \( \binom{n}{k} \), where \( n \) is the total power of the binomial and \( k \) is the specific term number you are looking at within the expansion. These coefficients are derived from a fascinating area of mathematics known as combinatorics, specifically related to combinations.

The Binomial Coefficient can be calculated using the formula:

• \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)

where \( n! \) (n factorial) is the product of all positive integers up to \( n \). This coefficient indicates how many ways you can choose \( k \) items from \( n \) items, and in the case of binomial expansion, it tells you the multiplier for each term in the expansion. In our example, \( \binom{9}{7} \) or equivalently \( \binom{9}{2} \) tells you how many sets of 7 negative bills and 2 positive bills determine the specific product in the expansion process.

Polynomial Algebra

Polynomial Algebra encompasses operations and expressions involving polynomials. A polynomial itself consists of variables, exponents, and coefficients, combined by addition, subtraction, and multiplication but never division by a variable. In the Binomial Theorem, a critical application of polynomial algebra comes to life.

When expanding a binomial expression, each resulting term is part of a larger polynomial. Each of these terms has its own coefficient, power of variables, and signs, all specified by the binomial formula. Mastery in polynomial algebra involves understanding how these terms interact. For example, in the expansion of \( (2a-b)^9 \), the terms follow from polynomial principles. Here, we manage the signs, coefficients, and variable exponents derived directly from the bracket powers.

The learning of polynomial algebra aids in seamless manipulation and simplification of these expanded terms, making complex algebraic expressions easy to handle and calculate.