Problem 38 - College Algebra

Question

For the following exercises, find the indicated term of each binomial without fully expanding the binomial. The fourth term of \(\left(x^{3}-\frac{1}{2}\right)^{10}\)

Step-by-Step Solution

Answer

The fourth term is \\(-15x^{21}\\).

Step 1: Understanding the Binomial Theorem

The Binomial Theorem gives us a way to expand expressions like \( (a + b)^n \) without writing out every term. It states that \( (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \), where \( \binom{n}{k} \) is a binomial coefficient.

Step 2: Identifying the Components

In our binomial expression \( (x^3 - \frac{1}{2})^{10} \), we have \( a = x^3 \), \( b = -\frac{1}{2} \), and \( n = 10 \). The fourth term corresponds to \( k = 3 \) because binomial terms start from \( k=0 \).

Step 3: Plugging into the Binomial Formula

The \(k^{th}\) term of \( (a+b)^n \) is given by \( \binom{n}{k} a^{n-k} b^k \). We need the fourth term, i.e., when \( k=3 \): \( \binom{10}{3} (x^3)^{10-3} \left(-\frac{1}{2}\right)^3 \).

Step 4: Calculating the Binomial Coefficient

Calculate \( \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \).

Step 5: Substituting into the Term Formula

Now compute each part: \( (x^3)^{7} = x^{21} \) and \( \left(-\frac{1}{2}\right)^3 = -\frac{1}{8} \). So, the fourth term is \( 120 imes x^{21} imes \left(-\frac{1}{8}\right) \).

Step 6: Simplifying the Expression

Simplify \( 120 \times -\frac{1}{8} = -15 \). Thus, the fourth term is \( -15x^{21} \).

Key Concepts

Binomial Coefficient

The binomial coefficient is like a special multiplier when we're dealing with binomial expansions. It's represented by \( \binom{n}{k} \), pronounced as "n choose k," and helps us find the specific terms in the expansion of a binomial expression. Imagine you have a deck of cards, and you want to know how many ways you can pick a few cards out. That's what the binomial coefficient does but with numbers.

To calculate \( \binom{n}{k} \), we use a formula involving factorials (which are products of consecutive numbers). The formula is:

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

In the exercise, we needed \( \binom{10}{3} \). This means picking 3 items out of 10 without caring about the order. We do this by simplifying \( \frac{10!}{3!7!} \). Simply put, it’s like dividing the ways to arrange 10 items by the ways to arrange the 3 picked and the remaining 7. For this, we got 120 after the calculations.

Remember, the binomial coefficient essentially tells you how many ways you can choose 'k' things from 'n' total things. Its application goes beyond just math to areas like statistics and computer science.

Polynomial Expansion

Polynomial expansion is a fundamental concept closely tied to the Binomial Theorem. It involves breaking down expressions like \((a + b)^n\) into a sum of terms. Each of these terms takes into account different powers of 'a' and 'b'. It's like unfolding a compact present into its full, visible form. This allows us to work with and analyze expressions that would otherwise be quite cumbersome.

Using the Binomial Theorem, the expression \((a + b)^n\) gets expanded into:

• \[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]

In the given problem, we have \((x^3 - \frac{1}{2})^{10}\). Applying the theorem doesn't require expanding it fully if we're only interested in certain terms. This makes the binomial theorem not just a formula but a powerful tool for simplifying complex algebraic expressions. Specifically, for the fourth term (\(k=3\)) in our polynomial, we use the components and multiply them out, skipping directly to the term of interest without expanding the entire series.

Polynomial expansion helps resolve complex problems into simpler, more manageable pieces which is crucial in both academic exercises and real-world calculations.

Algebraic Expressions

Algebraic expressions are the language of algebra, involving numbers, variables, and operators (like plus or minus). They are the building blocks that help us solve for unknowns and describe mathematical patterns. An expression can be simple, with just a single number or variable, or it could be a compound structure with numerous terms and operations.

The term “binomial” refers to an algebraic expression that consists of exactly two terms, like \((a + b)\). In the exercise, \((x^3 - \frac{1}{2})\) is a binomial. When raised to a power, as in \((x^3 - \frac{1}{2})^{10}\), it expands into a polynomial using the Binomial Theorem. Understanding these expressions and their expansions is crucial because they allow us to generalize and work with diverse mathematical problems.

Each part of the expression plays a role. Variables represent unknowns, coefficients scale the variables, and the overall structure dictates how terms interact. When solving for specific terms (like finding the fourth term in our current exercise), comprehension of these elements and their interplay is fundamental in algebra. This understanding not only helps in solving textbook problems but also in practical situations where predicting an outcome depends on multiple varying factors.