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!