If the remainder when 2x3 + 3x2 – 5kx + 15 is divided by x – k i

1.	If the remainder when 2x3 + 3x2 – 5kx + 15 is divided by x – k is 2k3, then find the value of k.1. ±√(15/2)2. ±√(5/2)3. ±√154. ±√2
Answer» Correct Answer - Option 1 : ±√(15/2) Given: P(x) = Dividend = 2x³ + 3x² – 5kx + 15 Divisor = x – k Remainder = 2k³ Calculation: As the remainder is 2k³ when divided by x – k For x – k = 0 , x = k P(k) = 2k³ 2k³ + 3k² – 5k² + 15 = 2k³ ⇒ –2k² + 15 = 0 ⇒ 2k² = 15 ⇒ k² = 15/2 ⇒ k = ±√(15/2) ∴ Value of k is ±√(15/2)

If the remainder when 2x3 + 3x2 – 5kx + 15 is divided by x – k is 2k3, then find the value of k.1. ±√(15/2)2. ±√(5/2)3. ±√154. ±√2

Answer» Correct Answer - Option 1 : ±√(15/2)

Given:

P(x) = Dividend = 2x³ + 3x² – 5kx + 15

Divisor = x – k

Remainder = 2k³

Calculation:

As the remainder is 2k³ when divided by x – k

For x – k = 0 , x = k

P(k) = 2k³

2k³ + 3k² – 5k² + 15 = 2k³

⇒ –2k² + 15 = 0

⇒ 2k² = 15

⇒ k² = 15/2

⇒ k = ±√(15/2)

∴ Value of k is ±√(15/2)

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If the remainder when 2x3 + 3x2 – 5kx + 15 is divided by x – k is 2k3, then find the value of k.1. ±√(15/2)2. ±√(5/2)3. ±√154. ±√2

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