To resolve \((2x + 1)^8\), you can expand it using the Binomial Theorem. The Binomial Theorem states that:

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

For \((2x + 1)^8\), \(a = 2x\), \(b = 1\), and \(n = 8\). The general term in the expansion will be:

\[
\binom{8}{k} (2x)^{8-k} \cdot 1^k = \binom{8}{k} \cdot (2x)^{8-k}
\]

Let's expand it:

\[
(2x + 1)^8 = \sum_{k=0}^{8} \binom{8}{k} \cdot (2x)^{8-k} \cdot 1^k
\]

Now let's calculate each term:

1. For \(k = 0\):
   \[
   \binom{8}{0} \cdot (2x)^8 = 1 \cdot 256x^8 = 256x^8
   \]
2. For \(k = 1\):
   \[
   \binom{8}{1} \cdot (2x)^7 = 8 \cdot 128x^7 = 1024x^7
   \]
3. For \(k = 2\):
   \[
   \binom{8}{2} \cdot (2x)^6 = 28 \cdot 64x^6 = 1792x^6
   \]
4. For \(k = 3\):
   \[
   \binom{8}{3} \cdot (2x)^5 = 56 \cdot 32x^5 = 1792x^5
   \]
5. For \(k = 4\):
   \[
   \binom{8}{4} \cdot (2x)^4 = 70 \cdot 16x^4 = 1120x^4
   \]
6. For \(k = 5\):
   \[
   \binom{8}{5} \cdot (2x)^3 = 56 \cdot 8x^3 = 448x^3
   \]
7. For \(k = 6\):
   \[
   \binom{8}{6} \cdot (2x)^2 = 28 \cdot 4x^2 = 112x^2
   \]
8. For \(k = 7\):
   \[
   \binom{8}{7} \cdot (2x)^1 = 8 \cdot 2x = 16x
   \]
9. For \(k = 8\):
   \[
   \binom{8}{8} \cdot (2x)^0 = 1 \cdot 1 = 1
   \]

Now combine all these terms together:

\[
(2x + 1)^8 = 256x^8 + 1024x^7 + 1792x^6 + 1792x^5 + 1120x^4 + 448x^3 + 112x^2 + 16x + 1
\]

Would you like a detailed explanation of any specific step?

---

1. How would you expand \((3x + 2)^6\)?
2. What is the Binomial Theorem and how is it applied?
3. How can Pascal's Triangle be used to expand binomials?
4. What is the coefficient of \(x^7\) in \((3x + 1)^9\)?
5. How would you apply the Binomial Theorem to \((x - 2)^5\)?

**Tip:** The coefficients in the expansion correspond to the entries in Pascal's Triangle.

Learn how to expand (2x + 1)^8 using the Binomial Theorem with a step-by-step guide. This problem covers key algebraic concepts, including the Binomial Theorem and combinatorics. Suitable for high school students in Grades 10-12.

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