Problem 6 - Algebra 2 and Trigonometry

Question

In \(3-10,\) write the expansion of each binomial. $$ (x+2)^{5} $$

Step-by-Step Solution

Answer

The expansion of \((x+2)^5\) is: \(x^5 + 10x^4 + 40x^3 + 80x^2 + 80x + 32\).

Step 1: Understanding the Binomial Theorem

The binomial theorem provides a way to expand expressions of the form \((a + b)^n\). For \((x + 2)^5\), it means we expand using the formula: \((a + b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}\).

Step 2: Identifying Parameters

In the expression \((x + 2)^5\), we identify \(a = x\), \(b = 2\), and \(n = 5\). We will use these to substitute into the binomial formula.

Step 3: Calculate Binomial Coefficients

For each term in the expansion, calculate the binomial coefficients \(\binom{5}{k}\), where \(k = 0, 1, 2, 3, 4, 5\). These coefficients are: \(\binom{5}{0} = 1\), \(\binom{5}{1} = 5\), \(\binom{5}{2} = 10\), \(\binom{5}{3} = 10\), \(\binom{5}{4} = 5\), \(\binom{5}{5} = 1\).

Step 4: Form Each Term

Use the binomial coefficients to form the terms of the expansion. Each term is given by \(\binom{5}{k}x^{5-k}2^{k}\). Calculate each term:- When \(k=0\): \(1 \cdot x^5 \cdot 2^0 = x^5\).- When \(k=1\): \(5 \cdot x^4 \cdot 2^1 = 10x^4\).- When \(k=2\): \(10 \cdot x^3 \cdot 2^2 = 40x^3\).- When \(k=3\): \(10 \cdot x^2 \cdot 2^3 = 80x^2\).- When \(k=4\): \(5 \cdot x^1 \cdot 2^4 = 80x\).- When \(k=5\): \(1 \cdot x^0 \cdot 2^5 = 32\).

Step 5: Write the Final Expansion

Combine all the terms obtained from Step 4 to write the complete expansion: \(x^5 + 10x^4 + 40x^3 + 80x^2 + 80x + 32\). Each term corresponds to the binomial coefficient multiplied by the powers of \(x\) and \(2\).

Key Concepts

Binomial Expansion Explained

The binomial expansion allows us to expand expressions that are raised to a power. Specifically, it helps us deal with expressions of the form \((a + b)^n\). In this expression, \(a\) and \(b\) are any numbers or symbols, and \(n\) is a whole number. The binomial theorem gives us a way to expand this expression into a series of terms.

The formula is:

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

Each term in the series is composed of binomial coefficients, powers of \(a\), and powers of \(b\). The theorem is incredibly useful for calculating powers of binomials without having to multiply the expression multiple times. It is a systematic way to find each term in the expansion.

Understanding Binomial Coefficients

Binomial coefficients serve as the numerical factors in the binomial expansion. They are denoted as \(\binom{n}{k}\), read as "n choose k," and are crucial in determining the weight or significance of each term in the expansion.

Binomial coefficients can be calculated using the formula:

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

Here, \(n!\) (n factorial) denotes the product of all positive integers up to \(n\), and similarly for \(k!\) and \((n-k)!\).

In practice, these coefficients correspond to entries in Pascal's Triangle, an easy visual reference for their values. For instance, if \(n = 5\), you would read across the sixth row of Pascal's Triangle to find the required coefficients for your expansion.

Diving into Polynomial Expansion

Polynomial expansion involves breaking down the expression \((x + 2)^5\) into a sum of terms with individual powers of \(x\) and constants. In our specific case, we use the binomial theorem to expand \((x + 2)^5\) as follows.

After applying the theorem, each term in the expansion \(x^5 + 10x^4 + 40x^3 + 80x^2 + 80x + 32\) is formed by calculating:

• A binomial coefficient, like \(\binom{5}{k}\).
• A power of \(x\), derived from \(x^{5-k}\).
• A power of 2, as in \((2)^k\).

By combining these parts, we systematically obtain each term. This process highlights how individual elements contribute to the full polynomial, revealing both structure and pattern beyond a seemingly complex expression.