When the polynomial P(x) = x^3 - 3x^2 + kx + 12 is divided by (x - 2), the remainder is 6.… — JAMB UTME

When the polynomial $P(x) = x^3 - 3x^2 + kx + 12$ is divided by $(x - 2)$ , the remainder is $6$ . What is the value of the constant $k$ ?

Answer: -1

The value of the constant

k

-1

By the Remainder Theorem, the remainder when

P(x)

is divided by

(x - 2)

P(2)

. Substituting

x = 2

into

P(x) = x^3 - 3x^2 + kx + 12

gives

P(2) = 8 - 12 + 2k + 12 = 8 + 2k

. Setting

8 + 2k = 6

yields

2k = -2

, so

k = -1

Apply the Remainder Theorem

P(2) = 6

Dividing

P(x)

(x - 2)

leaves a remainder equal to evaluating

P(x)

x = 2

Substitute

x = 2

into

P(x)

and set equal to

6

(2)^3 - 3(2)^2 + 2k + 12 = 6

Set the evaluated polynomial equal to the given remainder.

Simplify the arithmetic terms

8 + 2k = 6

Calculate powers and products:

8 - 12 + 12 = 8

Solve the linear equation for

k

k = -1

Subtract 8 from both sides to get

2k = -2

, then divide by 2.

Polynomial Remainder Theorem

Question