Problem 32 - College Algebra

Question

In Exercises 31-38, write the first three terms in each binomial expansion, expressing the result in simplified form. $$(x+3)^{6}$$

Step-by-Step Solution

Answer

The first three terms of the expansion of \((x+3)^6\) are \(x^{6}\), \(18*x^{5}\), and \(540*x^{4}\).

Step 1: Identify the components of the binomial theorem

For the given binomial \((x+3)^6\), a is x, b is 3, and n is 6. The first three terms will include when k equals 0, 1, and 2.

Step 2: Calculate the first term (k = 0)

Plugging a=x, b=3, n=6, and k=0 into the binomial theorem equation gives us: 6C0 * x^{6-0} * 3^0. Simplifying this gives \(x^{6}\) as the first term.

Step 3: Calculate the second term (k = 1)

Plugging a=x, b=3, n=6, and k=1 into the binomial theorem equation gives us: 6C1 * x^{6-1} * 3^1. Simplifying this gives \(18*x^{5}\) as the second term.

Step 4: Calculate the third term (k = 2)

Plugging a=x, b=3, n=6, and k=2 into the binomial theorem equation gives us: 6C2 * x^{6-2} * 3^2. Simplifying this gives \(540*x^{4}\) as the third term.

Key Concepts

Binomial Expansion

Binomial expansion refers to the process of expanding an expression that has been raised to a power, specifically when the expression is a binomial. A binomial is an algebraic expression with two terms, for instance, \(x + 3\). When we deal with binomial expansion, we essentially want to multiply this binomial by itself a certain number of times, which is indicative of the exponent.

Take the example from the exercise, \(x + 3)^6\). To manually expand this would require multiplying \(x + 3\) by itself six times, a tedious task to say the least. However, we can avoid the lengthy multiplication using the binomial theorem, which provides a formulaic method to find each term of the expansion directly.

Typically, the expanded form will follow a pattern governed by the coefficients known as binomial coefficients. These coefficients can be found using combinations, expressed as \(nCk\), which are also the entries of Pascal's Triangle. Recognizing these patterns and applying them appropriately can significantly simplify the process of binomial expansion.

Binomial Theorem

The binomial theorem is the backbone of binomial expansions. It provides a shortcut to expand expressions raised to a power without manually multiplying the binomial numerous times. The theorem states that for any positive integer \(n\), the expansion of a binomial \(a + b\) raised to the power of \(n\), denoted as \( (a + b)^n \), can be expressed as a sum of terms of the form \( nCk \cdot a^{n-k} \cdot b^k \), where \(k\) is an integer ranging from \(0\) to \(n\).

The factor \(nCk\) represents the binomial coefficient, which are the terms in Pascal's Triangle and is calculated as \(\frac{n!}{k!(n-k)!}\). This coefficient, combined with the appropriate powers of \(a\) and \(b\), gives us each term in the expansion. In practice, \(a\) and \(b\) can be any numbers or algebraic expressions, and power \(n\) can be any whole number, which makes the binomial theorem a versatile tool in algebra.

Simplified Form in Algebra

When we talk about a simplified form in algebra, we refer to the expression that is broken down to its most basic components without changing its value. Simplifying is a fundamental step to make expressions more understandable or to prepare them for further operations. For example, when we expand a binomial expression, each term usually contains a number of multiplications that can often be carried out to achieve a more concise form.

In the given exercise, the terms \(6C0 \cdot x^{6-0} \cdot 3^0\), \(6C1 \cdot x^{6-1} \cdot 3^1\), and \(6C2 \cdot x^{6-2} \cdot 3^2\) are simplified to \(x^6\), \(18x^5\), and \(540x^4\) respectively. These simplified terms are much easier to work with and far more approachable when compared to their unsimplified counterparts. Moreover, they are crucial when the goal is to perform additional algebraic operations like differentiation or integration, or to solve for certain values of \(x\). The simplified form is essentially algebra at its cleanest and most efficient state.