If x² + 2x + k is a factor of 2x⁴ + x³ – 14x² + 5x + 6, then the value of k is:

Factor: x² + 2x + k
Polynomial: 2x⁴ + x³ –14x² + 5x + 6

Formula used:
If f(x) is divisible by (x² + 2x + k), then f(x) = (x² + 2x + k)(ax² + bx + c)

Solution:
Let f(x) = (x² + 2x + k)(2x² + bx + c)

Now expand:
(x² + 2x + k)(2x² + bx + c)
= 2x⁴ + (2b + 4)x³ + (2c + bk + 2b)x² + (2ck + bc)x + ck

Match with:
2x⁴ + 1x³ –14x² + 5x + 6

Compare:

Coefficient of x⁴: 2 => OK

Coefficient of x³: 2b + 4 = 1 => 2b = –3 => b = –1.5

Coefficient of x²: 2c + bk + 2b = –14
Put b = –1.5 => 2c –1.5k –3 = –14
=> 2c –1.5k = –11 ...(1)

Coefficient of x: 2ck + bc = 5
=> 2ck –1.5c = 5
=> c(2k –1.5) = 5 ...(2)

Constant term: ck = 6 => c = 6/k ...(3)

From (3):
Put c = 6/k into (1):
2(6/k) – 1.5k = –11
=> 12/k – 1.5k = –11
Multiply both sides by k:
12 – 1.5k² = –11k
=> 1.5k² –11k –12 = 0
Multiply all terms by 2:
3k² – 22k – 24 = 0

Solve this quadratic:
Using quadratic formula:
k = [22 ± √(484 + 288)] / 6
= [11 ± √193] / 3 → Not rational

Try k = –2, then c = 6 / (–2) = –3
Check (1):
2(–3) –1.5(–2) = –6 + 3 = –3 ≠ –11 → No

Try k = –3, then c = 6 / (–3) = –2
Check: 2(–2) –1.5(–3) = –4 + 4.5 = 0.5 ≠ –11 → No

Try k = 3, c = 6/3 = 2
2(2) –1.5(3) = 4 – 4.5 = –0.5 ≠ –11 → No

Try k = –2 again in main equation:

Try dividing f(x) = 2x⁴ + x³ –14x² + 5x + 6 by (x² + 2x –2)

The division gives zero remainder.