What is the value of k for which when x^3 - 2x^2 + 3kx + 18 is divided by (x - 6), the remainder is equal to zero?

You should write the factored form of polynomial such
that:

`x^3 - 2x^2 + 3kx + 18 = (x - 6)(ax^2 + bx +
c)`

You need to open the brackets to the right side such
that:

`x^3 - 2x^2 + 3kx + 18 = ax^3 + bx^2 + cx - 6ax^2 -
6bx - 6c`

You need to collect like terms to the right side
such that:

`x^3 - 2x^2 + 3kx + 18 = ax^3 + x^2(b - 6a) +
x(c - 6b) - 6c`

Equating coefficients of like powers
yields:

`a = 1`

b - 6a = -2
=> b - 6 = -2 => b = 4

`c - 6b = 3k =gt c -
24 = 3k =gt c = 3k + 24`

`-6c = 18 =gt c =
-3`

You need to substitute -3 for c in `c = 3k + 24`  such
that:

`-3 = 3k + 24`

Dividing
by 3 both sides yields:

`-1 = k + 8 =gt k =
-9`

Hence, evaluating k using th factored
form of polynomial yields `k = -9` .