The sum of the remainders obtained when polynomial $2x^3 + (p+2)x + (p-2)$ is divided by (x – 2) and when it is divided by (x + 1) is 0. Then the value of p is :

Q: The sum of the remainders obtained when polynomial $2x^3 + (p+2)x + (p-2)$ is divided by (x – 2) and when it is divided by (x + 1) is 0. Then the value of p is :

To determine the value of $ p $ such that the sum of the remainders when the polynomial $ 2x^3 + (p+2)x + (p-2) $ is divided by $ (x - 2) $ and $ (x + 1) $ is zero, we need to evaluate the polynomial at the roots of these divisors. ### Step-by-Step Explanation: 1. **Remainder when divided by $ (x - 2) $:** By the Remainder Theorem, the remainder of a polynomial $ f(x) $ when divided by $ (x - a) $ is $ f(a) $. \[ f(2) = 2(2)^3 + (p+2)(2) + (p-2) \] \[ = 2 \times 8 + 2(p+2) + (p-2) \] \[ = 16 + 2p + 4 + p - 2 \] \[ = 18 + 3p \] 2. **Remainder when divided by $ (x + 1) $:** Similarly, the remainder when divided by $ (x + 1) $ is $ f(-1) $. \[ f(-1) = 2(-1)^3 + (p+2)(-1) + (p-2) \] \[ = 2(-1) - (p+2) + (p-2) \] \[ = -2 - p - 2 + p - 2 \] \[ = -6 \] 3. **Sum of the remainders:** According to the problem, the sum of these remainders is zero. \[ (18 + 3p) + (-6) = 0 \] \[ 18 + 3p - 6 = 0 \] \[ 12 + 3p = 0 \] \[ 3p = -12 \] \[ p = -4 \] Thus, the value of $ p $ that satisfies the condition is $-4$, which corresponds to **Option 3**. ### Explanation for Incorrect Options: - **Option 1 (p = 3):** If $ p = 3 $, the sum of the remainders would not be zero. Substituting $ p = 3 $ into the equations would yield a non-zero sum, contradicting the problem's condition. - **Option 2 (p = -2):** Similarly, substituting $ p = -2 $ would result in a sum that is not zero, failing to meet the requirement. - **Option 4 (p = 6):** If $ p = 6 $, the calculations would again lead to a non-zero sum, which does not satisfy the condition given in the problem. Therefore, the only value of $ p $ that makes the sum of the remainders zero is $-4$, confirming that **Option 3** is indeed the correct answer.

Question

Accepted Answer

To determine the value of $ p $ such that the sum of the remainders when the polynomial $ 2x^3 + (p+2)x + (p-2) $ is divided by $ (x - 2) $ and $ (x + 1) $ is zero, we need to evaluate the polynomial at the roots of these divisors.

### Step-by-Step Explanation:

1. **Remainder when divided by $ (x - 2) $:**

By the Remainder Theorem, the remainder of a polynomial $ f(x) $ when divided by $ (x - a) $ is $ f(a) $.

$$
   f(2) = 2(2)^3 + (p+2)(2) + (p-2)
   $$
   $$
   = 2 \times 8 + 2(p+2) + (p-2)
   $$
   $$
   = 16 + 2p + 4 + p - 2
   $$
   $$
   = 18 + 3p
   $$

2. **Remainder when divided by $ (x + 1) $:**

Similarly, the remainder when divided by $ (x + 1) $ is $ f(-1) $.

$$
   f(-1) = 2(-1)^3 + (p+2)(-1) + (p-2)
   $$
   $$
   = 2(-1) - (p+2) + (p-2)
   $$
   $$
   = -2 - p - 2 + p - 2
   $$
   $$
   = -6
   $$

3. **Sum of the remainders:**

According to the problem, the sum of these remainders is zero.

$$
   (18 + 3p) + (-6) = 0
   $$
   $$
   18 + 3p - 6 = 0
   $$
   $$
   12 + 3p = 0
   $$
   $$
   3p = -12
   $$
   $$
   p = -4
   $$

Thus, the value of $ p $ that satisfies the condition is $-4$, which corresponds to **Option 3**.

### Explanation for Incorrect Options:

- **Option 1 (p = 3):** 
  If $ p = 3 $, the sum of the remainders would not be zero. Substituting $ p = 3 $ into the equations would yield a non-zero sum, contradicting the problem's condition.

- **Option 2 (p = -2):**
  Similarly, substituting $ p = -2 $ would result in a sum that is not zero, failing to meet the requirement.

- **Option 4 (p = 6):**
  If $ p = 6 $, the calculations would again lead to a non-zero sum, which does not satisfy the condition given in the problem.

Therefore, the only value of $ p $ that makes the sum of the remainders zero is $-4$, confirming that **Option 3** is indeed the correct answer.