In the expansion of the expression below, the constant term is 16 128. Find the value(s) of k.

Question

In the expansion of the expression below, the constant term is 16 128. Find the value(s) of k. I don't understand how to do this, if someone could provide help it would greatly be appriciated

x²(k/x + 3x²)⁸

Answer 1

k=2

the constant term of the expansion is the 3rd term of the 8th row of Pascals Triangle: 1, 8, 28, ...

28(k^6/x^6)(3x^2)^2

=256k^6/x^2

multiply that by x^2 gives the constant term 256k^6 = 16128

k^6 = 16128/256 = 64

k= 6th root of 64 = 2

2^6 = 64

the general formula for a binomial expansion

is

(a+b)^n = a^n + nC1(a^n-1)(b^1) + nC2(a^n-2)(b^2) + ....nCr(a^n-r)(b^r) + .... b^8

in this specific example, a = k/x and b = 3x^2 n=8, 8C2 = 28

only that 3rd term when multiplied by x^2 will give you a constant term

plug the numbers in and you get

28((k/x)^6)(3x^2)^2 simplify and set equal to 16128

It helps to review the binomial expansion

(a+b)^2 = a^2 + 2ab + b^2

(a+b)^3 = a^3 +3a^2b +3ab^2 + b^3

...

(a+b)^8 = a^8 + 8a^7b + 28a^6b^2 + ...

It's that 3rd term that combines with x^2 to get a product that's a constant term, given the values of a and b in the problem given