Problem 24 - College Algebra

Question

Use the Binomial Theorem to expand each binomial and express the result in simplified form. $$ (c+3)^{5} $$

Step-by-Step Solution

Answer

The expanded form of \((c+3)^{5}\) is \(c^{5}+15c^{4}+90c^{3}+270c^{2}+405c+243\).

Step 1: Identify the binomial and the power

Here, \(c+3\) is the binomial and it is powered to 5.

Step 2: Apply Binomial Theorem formula

The Binomial Theorem can be represented as: \( (x+y)^{n}=\sum_{k=0}^{n} {n\choose k}x^{n-k}y^{k}\). Thus, apply this formula by replacing \(x\) with \(c\), \(y\) with \(3\), and \(n\) with \(5\). This would result in five terms. The 'n choose k' (\({n\choose k}\)) term denotes the binomial coefficient.

Step 3: Calculate each term

Calculate each of the five terms: \({5\choose 0}c^{5}*3^{0}\), \({5\choose 1}c^{4}*3^{1}\), \({5\choose 2}c^{3}*3^{2}\), \({5\choose 3}c^{2}*3^{3}\), \({5\choose 4}c^{1}*3^{4}\), and \({5\choose 5}c^{0}*3^{5}\). Here, \({n\choose k}\), is the binomial coefficient which can be calculated as \(\frac{n!}{k!(n-k)!}\), where \(!\) denotes the factorial.

Step 4: Simplify the terms

After expanding each term, simplify it. The results will be: \(c^{5}\), \(15c^{4}\), \(90c^{3}\), \(270c^{2}\), \(405c\), and \(243\).

Step 5: Compose the final solution

Combine resulting terms to form the expanded binomial: \(c^{5}+15c^{4}+90c^{3}+270c^{2}+405c+243\).

Key Concepts

Binomial Coefficient

The binomial coefficient, denoted as \({n\choose k}\), plays a crucial role in the expansion of binomials using the Binomial Theorem. It determines the number of ways to choose \(k\) elements from a set of \(n\) elements, which is pivotal when calculating the terms in a binomial expansion.

To compute the binomial coefficient, you can use the formula:

• \[{n\choose k} = \frac{n!}{k!(n-k)!}\]

Here, the exclamation mark (\(!\)) represents a factorial, which is the product of an integer and all the integers below it.

For example, when expanding \((c+3)^5\) using the Binomial Theorem, the terms include coefficients like \({5\choose 2}\) which equals \(\frac{5!}{2!(5-2)!} = 10\). This makes the coefficient 10 for that specific term.

Factorial

Factorial is a mathematical operation, represented by the symbol \(!\). It involves multiplying a number by all the positive integers less than itself. For instance, the factorial of 5 (written as \(5!\)) is:

• \[5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\]

Factorials are essential in calculating the binomial coefficients in the Binomial Theorem. For example, in the expansion of \((c+3)^5\), calculating \({5\choose 3}\) involves finding \(5!\), \(3!\), and \((5-3)!\). The ability to compute factorials accurately ensures you can determine the correct coefficients for each term.

Polynomial Expansion

Polynomial expansion refers to the process of expressing a power of a binomial (like \((c+3)^5\)) as a sum of terms raised to a power. The Binomial Theorem aids in expanding these expressions efficiently, without multiplying the binomial multiple times.

Using the theorem, we express \((c+3)^5\) as a series of terms:

• \(c^5\)
• \(15c^4\)
• \(90c^3\)
• \(270c^2\)
• \(405c\)
• \(243\)

The method involves computing each term by using the binomial coefficient and the values of \(x\) and \(y\) from the binomial. Each term corresponds to an increasing power of the second term \(y\), and a decreasing power of the first term \(x\), aligning with the binomial structure.

Algebraic Expression

An algebraic expression consists of numbers, variables, and arithmetic operations. In the context of the Binomial Theorem, the expression \((c+3)^5\) is a binomial algebraic expression.

The goal of using the Binomial Theorem is to express this in an expanded algebraic form with terms systematically derived. Each part of the algebraic expression in the expansion consists of:

• Binomial coefficient (computed from the factorial)
• Variable \(c\) raised to a power
• Constant number 3 raised to a power
• Product of these components

After expansion, algebra captures the structure of a polynomial, transforming it to standard form: \(c^5 + 15c^4 + 90c^3 + 270c^2 + 405c + 243\), facilitating further algebraic manipulations or evaluations.