Find the coefficient of x^7 in the expansion of (2x-3)^7
Pick the correct answer
-2187, -128, 2, or 128

To find the coefficient of \( x^7 \) in the expansion of \( (2x - 3)^7 \), we can use the **Binomial Theorem**. The Binomial Theorem states:

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

In this case, \( a = 2x \), \( b = -3 \), and \( n = 7 \).

We want the term where the power of \( x \) is exactly 7. In the expansion, the general term looks like this:

\[
T_k = \binom{7}{k} (2x)^{7-k} (-3)^k
\]

We need the power of \( x \) to be 7, which means the exponent of \( x \) in \( (2x)^{7-k} \) must be 7. This occurs when \( 7 - k = 7 \), or \( k = 0 \).

Now, let's calculate the term when \( k = 0 \):

\[
T_0 = \binom{7}{0} (2x)^7 (-3)^0
\]

This simplifies to:

\[
T_0 = 1 \cdot (2x)^7 \cdot 1 = (2x)^7 = 128x^7
\]

Thus, the coefficient of \( x^7 \) is \( 128 \).

The correct answer is **128**.

Would you like more details on this process, or do you have any questions?

Here are 5 related questions for further exploration:
1. How would the coefficient change if the exponent were higher, say \( (2x - 3)^{10} \)?
2. What if the signs inside the binomial were different, such as \( (2x + 3)^7 \)?
3. How can the Binomial Theorem be applied to expansions with fractional exponents?
4. How would you expand \( (2x - 3)^7 \) without using the Binomial Theorem?
5. What is the general form of the Binomial Theorem for negative exponents?

**Tip**: Always double-check the signs in binomial expansions, especially when terms involve subtraction!

Find the coefficient of x^7 in the expansion of (2x - 3)^7. Pick the correct answer: -2187, -128, 2, or 128.

Discover how to find the coefficient of x^7 in the expansion of (2x - 3)^7 using the Binomial Theorem. This algebra problem highlights key concepts in binomial expansion and exponents, suitable for students in Grades 9-12. The correct answer is 128.

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