Find the coefficient of ๐‘ฅ4

in the expansion of
(1 โˆ’ 2๐‘ฅ)6.
A. -320
B. -240
C. 240
D. 320

Using the binomial theorem, we have:

(1 - 2๐‘ฅ)6 = 1 - 6(2๐‘ฅ) + 15(2๐‘ฅ)2 - 20(2๐‘ฅ)3 + 15(2๐‘ฅ)4 - 6(2๐‘ฅ)5 + (2๐‘ฅ)6

The coefficient of ๐‘ฅ4 is the coefficient of the term 15(2๐‘ฅ)4, which is 15(16) = 240. Therefore, the answer is C.

To find the coefficient of ๐‘ฅ^4 in the expansion of (1 - 2๐‘ฅ)^6, we can use the binomial theorem.

The binomial theorem states that for any positive integer n:

(๐‘Ž + ๐‘)^๐‘› = ๐ถ(๐‘›, 0)๐‘Ž^๐‘› + ๐ถ(๐‘›, 1)๐‘Ž^(๐‘›โˆ’1)๐‘ + ๐ถ(๐‘›, 2)๐‘Ž^(๐‘›โˆ’2)๐‘^2 + ... + ๐ถ(๐‘›, ๐‘›โˆ’1)๐‘Ž๐‘^(๐‘›โˆ’1) + ๐ถ(๐‘›, ๐‘›)๐‘^๐‘›,

where ๐ถ(๐‘›, ๐‘Ÿ) is the binomial coefficient, which is calculated as:

๐ถ(๐‘›, ๐‘Ÿ) = ๐‘›! / (๐‘Ÿ! (๐‘›โˆ’๐‘Ÿ)!)

In this case, ๐‘› = 6 and ๐‘Ž = 1, ๐‘ = -2๐‘ฅ. We are interested in the term with ๐‘ฅ^4, so ๐‘Ÿ = 4.

๐ถ(6, 4) = 6! / (4! (6โˆ’4)!) = 6! / (4! 2!) = (6 * 5 * 4 * 3 * 2 * 1) / (4 * 3 * 2 * 1 * 2 * 1) = (6 * 5) / (2 * 1) = 15

So the coefficient of ๐‘ฅ^4 in the expansion is 15. However, we need to consider the coefficient of ๐‘ฅ^4 multiplied by -2^4 = -16.

Therefore, the coefficient of ๐‘ฅ^4 in the expansion of (1 - 2๐‘ฅ)^6 is -16 * 15 = -240.

Therefore, the correct answer is B. -240.