p(x) be a polynomial of degree at most 5 which leaves remainder - 1 and 1 upon division by `(x-1)^3` and `(x+1)^3`respectively, the number of real roots of `P(x) = 0` is

Options

A. `-(5)/(3)`
B. `-(10)/(3)`
C. `2`
D. `-5`

Explanation

To solve the problem, we need to find the number of real roots of the polynomial \( P(x) \) given the conditions on its remainders when divided by \( (x-1)^3 \) and \( (x+1)^3 \).

### Step-by-Step Solution:

1. **Understanding the Polynomial**:
Since \( P(x) \) is a polynomial of degree at most 5, we can express it as:
\[ P(x) = ax^5 + bx^3 + cx + d \]
Here, we note that \( P(x) \) is an odd function due to the nature of the remainders.

2. **Using the Remainder Conditions**:
- From the condition \( P(1) = -1 \):
\[ a(1)^5 + b(1)^3 + c(1) + d = -1 \implies a + b + c + d = -1 \quad \text{(Equation 1)} \]
- From the condition \( P(-1) = 1 \):
\[ a(-1)^5 + b(-1)^3 + c(-1) + d = 1 \implies -a - b - c + d = 1 \quad \text{(Equation 2)} \]

3. **Setting Up the System of Equations**:
We now have two equations:
- \( a + b + c + d = -1 \) (Equation 1)
- \( -a - b - c + d = 1 \) (Equation 2)

4. **Solving the System**:
We can add Equation 1 and Equation 2:
\[ (a + b + c + d) + (-a - b - c + d) = -1 + 1 \]
This simplifies to:
\[ 2d = 0 \implies d = 0 \]
Substituting \( d = 0 \) back into Equation 1:
\[ a + b + c = -1 \quad \text{(Equation 3)} \]

5. **Finding the Derivatives**:
We differentiate \( P(x) \):
\[ P'(x) = 5ax^4 + 3bx^2 + c \]
We also need the second derivative:
\[ P''(x) = 20ax^3 + 6bx \]

6. **Analyzing the Roots**:
Since \( P(x) \) is an odd function, it will have symmetry about the origin. This means if \( P(x) \) has a root at \( x = r \), it will also have a root at \( x = -r \).

7. **Finding the Critical Points**:
We need to find the critical points by solving \( P'(x) = 0 \):
\[ 5ax^4 + 3bx^2 + c = 0 \]
This is a quadratic in \( x^2 \). The number of real roots of \( P'(x) = 0 \) will determine the number of turning points.

8. **Determining the Number of Roots**:
Since \( P(x) \) is of degree 5, it can have at most 5 real roots. Given that \( P(x) \) is an odd function and we have established that it has one real root at \( x = 0 \), we need to check the behavior of \( P(x) \) around this point.

9. **Conclusion**:
Since \( P(x) \) is a continuous function and changes sign around \( x = 0 \), and given that it is monotonically decreasing (as derived from the negative value of \( P'(0) \)), we conclude that \( P(x) = 0 \) has exactly **one real root**.

### Final Answer:
The number of real roots of \( P(x) = 0 \) is **1**.