# 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?

sciencesolve | Certified Educator

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` .

justaguide | Certified Educator

If the remainder when the function f(x) = x^3 - 2*x^2 + 3*k*x + 18 is divided by (x - 6) has to be equal to zero, the value of k can be derived by the remainder theorem.

The remainder when f(x) is divided by (x - 6) is f(6).

f(6) = 0

=> 6^3 - 2*6^2 + 3*k*6 + 18 = 0

=> 216 - 2*36 + 3*k*6 + 18 = 0

=> 216 - 72 + 18k + 18 = 0

=> 162 + 18k = 0

=> 18*k = -162

=> k = -162/18

=> k = -9

If the remainder we get when x^3 - 2*x^2 + 3*k*x + 18 is divided by (x - 6) is 0 the value of k = -9.

giorgiana1976 | Student
To determine the value of k we'll apply the reminder theorem. We know that the polynomial P(x)= x^3 - 2x^2 + 3kx + 18 at x=6 is the same as the reminder we'll get when we divide P(x) by x-6.

Since the reminder is 0, then we'll get P(6)=0

Therefore, we'll calculate P(6) to determine the value of k:

P(6) = 6^3 - 2*6^2 + 3*k*6 + 18

P(6) = 216 - 72 + 18k + 18 But P(6) = 0 => 162+18k = 0

18k = -162

k = -162/18

k=-9

The requested value of k is: k = -9.