Problem 39 - A Graphical Approach to Precalculus with Limits

Question

Write the indicated tem of each binomial expansion. Sixth term of \((4 h-j)^{8}\)

Step-by-Step Solution

Answer

The sixth term of \((4h-j)^8\) is \(-3584h^3j^5\).

Step 1: Identify the Formula for Binomial Expansion

The binomial expansion of \((a + b)^n\) is given by: \[\sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\] where \( \binom{n}{k} \) is the binomial coefficient. For each term, we use the formula: \[T_{k+1} = \binom{n}{k} a^{n-k} b^k\]

Step 2: Identify the Values for the Variables

In the given binomial \((4h - j)^{8}\), we have:- \(a = 4h\)- \(b = -j\)- \(n = 8\) We are finding the sixth term, so \(k = 5\) (since the term corresponds to \(k+1\) in the formula).

Step 3: Calculate the Binomial Coefficient

The binomial coefficient is \(\binom{8}{5}\). Calculate it using the formula:\[\binom{8}{5} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56\]

Step 4: Compute the Powers of Terms

According to the formula for the term:- The power for \(a = (4h)\) is \(n-k = 8-5 = 3\)- The power for \(b = (-j)\) is \(k = 5\)Calculate these powers:\((4h)^3 = (4)^3 (h)^3 = 64h^3\)\((-j)^5 = -j^5\)

Step 5: Assemble the Term

Using the formula \(T_{k+1} = \binom{n}{k} a^{n-k} b^k\), substitute the calculated values:\[T_6 = 56 \times 64h^3 \times (-j)^5\]Multiply these to get:\(T_6 = 56 \times 64 \times h^3 \times (-j)^5 = -3584h^3j^5\)

Step 6: Simplify the Result

The sixth term of the expansion \((4h - j)^8\) simplifies to: \(-3584h^3j^5\). Therefore, the answer does not require further simplification.

Key Concepts

Binomial Coefficient

In the context of a binomial expansion, the binomial coefficient plays a crucial role. Think of it as a special number that tells us how many different ways we can pick items from a group. It's represented by \( \binom{n}{k} \), pronounced as "n choose k." This coefficient is essential in calculating each term in the expansion of \((a + b)^n\). To compute the binomial coefficient, you use the formula:

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

Where \( n! \) (pronounced "n factorial") is the product of all positive integers less than or equal to \( n \).
For example, in the expansion of \((4h - j)^8\), when calculating the sixth term, we identified that \( k = 5 \), and thus calculated:

• \[ \binom{8}{5} = 56 \]

It's crucial to get these coefficients right as they weigh the impact of each term's contribution to the final polynomial.

Polynomial Terms

Polynomial terms are expressions of the form \(a^{n-k} b^k\) in the binomial expansion. They involve not just the coefficients but also the powers of the expressions involved. Each term in a polynomial combines multiple factors like coefficients and variables raised to a power.
This helps form the expanded expression clearly. In our example of \((4h - j)^8\), we tackled the problem by specifying these terms for the sixth position in the sequence:

• \((4h)^3 = 64h^3\)
• \((-j)^5 = -j^5\)

By placing these powers together, each polynomial term comes to life.
This method demonstrates how the interplay of binomial coefficients and polynomial terms forms each distinct part of the expanded equation.

Exponents

Exponents describe how many times a number or a variable is multiplied by itself. In terms of binomial expansion, the role of exponents is to represent the power each variable is taken to. This power defines the dominance of each component within the binomial raised to a power. In simpler terms, it tells you how significant that piece is within the binomial context.
To identify them for each term in the expansion of \((a + b)^n\), use the breakdown from the formula \(a^{n-k} b^k\). In our example, for the sixth term calculated as:

• The exponent for \( (4h) \) was \( 3 \) given by \( n-k = 8-5 \)
• The exponent for \(-j\) was \( 5 \) resulting directly from \( k = 5 \)

These exponents determine the specific contribution and balance of each term.
IA better understanding of exponents is pivotal. It's the accuracy in dictating powers that leads to the correct expansion.