Question

if ×3+6×2+k×+6 is exactly divisible by (×+2), then k=?
1) -6
2) -7
3) -8
4) 11​

Answer 1

$x + 2 = 0 \\ x = - 2$

${x}^{3} + 6 {x}^{2} + kx + 6 \\ - 8 + 2 - 2k + 6 = 0 \\ - 8 + 8 - 2k = 0 \\ - 2k = 0 \\ k = 0$

Answer 2

Answer:

11

Step-by-step explanation:

refer attachment

hope this helps you

LAST STEPS (IF NOT CLEAR) OF 2ND METHOD

(k-8)x- 3x =0

(k-8)x=3x

x on both sides gets cancelled

k-8 =3

k =8+3

k=11