Find the coefficient of x^3 in the expansion of (2-x)^5

To find the coefficient of \(x^3\) in the expansion of \((2 - x)^5\), we will use the **Binomial Theorem**, which states:

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

Here, \(a = 2\), \(b = -x\), and \(n = 5\). The general term in the binomial expansion is:

\[
T_k = \binom{n}{k} a^{n-k} b^k
\]

Substituting into the formula:

\[
T_k = \binom{5}{k} (2)^{5-k} (-x)^k
\]

We are interested in the term that contains \(x^3\). This means \(k = 3\), so we substitute \(k = 3\) into the general term:

\[
T_3 = \binom{5}{3} (2)^{5-3} (-x)^3
\]

Now, calculate each part:

\[
\binom{5}{3} = \frac{5 \times 4}{2 \times 1} = 10
\]
\[
(2)^{5-3} = 2^2 = 4
\]
\[
(-x)^3 = -x^3
\]

Now, combine these:

\[
T_3 = 10 \times 4 \times (-x^3) = -40x^3
\]

Thus, the coefficient of \(x^3\) is \(-40\).

Would you like any further details or clarification?

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Here are some related questions to explore:
1. What is the general term in the binomial expansion of \((a - b)^n\)?
2. How would you find the coefficient of \(x^2\) in the same expansion?
3. Can you find the constant term in the expansion of \((2 - x)^5\)?
4. How does the sign of terms alternate in binomial expansions involving negative terms?
5. How does Pascal's Triangle relate to binomial coefficients?

**Tip:** When expanding binomials, always check the signs of terms, especially when dealing with negative numbers!

Find the coefficient of x^3 in the expansion of (2 - x)^5

Discover how to find the coefficient of x^3 in the expansion of (2 - x)^5 using the Binomial Theorem. This problem involves algebraic binomial expansion techniques and is suitable for high school students in Grades 9-12.

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