The Value of 'k' for which 2x²+3x³+2kx²- 3x+6 is exactly divisible

1.	The Value of 'k' for which 2x²+3x³+2kx²- 3x+6 is exactly divisible by (X+2)?
Answer» Answer: k = 1/2 Step-by-step explanation: suppose 2x²+3x³+2kx²-3x+6 = f(x) x+2 = 0 => x = -2 if 2x²+3x³+2kx²-3x+6 is exactly divisible by (x+2), f(-2) = 0 f(-2) = (2(-2)²)+(3(-2)³)+(2k(-2)²)-(3(-2))+6 = (24)+(3(-8))+(2k*4)-(-6)+6 = 8+(-24)+8k+6+6 = 20+8k-24 = 8k-4 f(-2) = 0 8k-4 = 0 => 8k = 4 => k = 4/8 => k = 1/2

The Value of 'k' for which 2x²+3x³+2kx²- 3x+6 is exactly divisible by (X+2)?

Answer»

Answer:

k = 1/2

Step-by-step explanation:

suppose 2x²+3x³+2kx²-3x+6 = f(x)

x+2 = 0

=> x = -2

if 2x²+3x³+2kx²-3x+6 is exactly divisible by (x+2),

f(-2) = 0

f(-2) = (2*(-2)²)+(3*(-2)³)+(2*k*(-2)²)-(3*(-2))+6

= (2*4)+(3*(-8))+(2*k*4)-(-6)+6

= 8+(-24)+8k+6+6

= 20+8k-24

= 8k-4

f(-2) = 0

8k-4 = 0

=> 8k = 4

=> k = 4/8

=> k = 1/2