(x+4)^5

Jasonbradley

4 min read

·

Jun 16, 2024

7

Share

Expanding (x+4)^5: A Journey Through the Binomial Theorem

The expression (x+4)^5 might look intimidating at first glance, but fear not! We can conquer this seemingly complex expansion with the help of the Binomial Theorem.

Understanding the Binomial Theorem

The Binomial Theorem provides a powerful formula to expand any expression of the form (x + y)^n, where n is a non-negative integer. The theorem states:

(x + y)^n = ∑_(k=0)^n (n choose k) * x^(n-k) * y^k

Let's break down this formula:

• (n choose k), also written as nCk or n_C_k, represents the binomial coefficient, calculated as n! / (k! * (n-k)!). This tells us how many ways we can choose k items from a set of n items.
• ∑_(k=0)^n indicates we sum over all values of k from 0 to n.
• x^(n-k) * y^k represents the terms of the expansion, with the powers of x decreasing and the powers of y increasing.

Applying the Theorem to (x+4)^5

Now, let's apply the Binomial Theorem to our expression (x+4)^5:

1. Identify n: In our case, n = 5.
2. Expand the sum: We'll need to sum the terms for k = 0, 1, 2, 3, 4, and 5.

Let's calculate each term:

• k = 0: (5 choose 0) * x^(5-0) * 4^0 = 1 * x^5 * 1 = x^5
• k = 1: (5 choose 1) * x^(5-1) * 4^1 = 5 * x^4 * 4 = 20x^4
• k = 2: (5 choose 2) * x^(5-2) * 4^2 = 10 * x^3 * 16 = 160x^3
• k = 3: (5 choose 3) * x^(5-3) * 4^3 = 10 * x^2 * 64 = 640x^2
• k = 4: (5 choose 4) * x^(5-4) * 4^4 = 5 * x^1 * 256 = 1280x
• k = 5: (5 choose 5) * x^(5-5) * 4^5 = 1 * x^0 * 1024 = 1024

Finally, combining all the terms, we get the expanded form of (x+4)^5:

(x+4)^5 = x^5 + 20x^4 + 160x^3 + 640x^2 + 1280x + 1024

Conclusion

The Binomial Theorem provides a systematic and efficient method for expanding expressions of the form (x + y)^n. By understanding and applying this theorem, we can conquer seemingly complex expressions like (x+4)^5 and gain a deeper understanding of polynomial expansions.