Answer:
The coefficient of x^2 in the expansion of (4+kx)(2-x)^6 is given by the product of the coefficients of x^2 in the two factors.
The coefficient of x^2 in (4+kx) is k times the coefficient of x in (4+kx), which is 4k.
The coefficient of x^2 in (2-x)^6 is given by the binomial coefficient (6,2) times (2)^4 times (-x)^2, which is 15*16x^2 or 240x^2.
Therefore, we have:
4k*240 = 384
Solving for k, we get:
k = 8/5
To find the coefficient of x^3 in the expansion, we use the same method.
The coefficient of x^3 in (4+kx)(2-x)^6 is given by the sum of the products of the coefficients of x^3 in the two factors.
The coefficient of x^3 in (4+kx) is k times the coefficient of x^3 in (4+kx), which is 4k.
The coefficient of x^3 in (2-x)^6 is given by the binomial coefficient (6,3) times (2)^3 times (-x)^3, which is -160x^3.
Therefore, we have:
4k*(-160) = -640k
Substituting k = 8/5, we get:
-640(8/5) = -1024
Therefore, the coefficient of x^3 in the expansion is -1024.