The number of polynomials of the form `x^(3)+ax^(2)+bx+c` that are divisible by `x^(2)+1`, where `a, b,cin{1,2,3,4,5,6,7,8,9,10}`, is

Options

A. 10
B. 15
C. 5
D. 8

Explanation

To solve the problem of finding the number of polynomials of the form \(x^3 + ax^2 + bx + c\) that are divisible by \(x^2 + 1\), where \(a, b, c \in \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Divisibility Condition**:
A polynomial \(P(x)\) is divisible by \(x^2 + 1\) if the remainder when \(P(x)\) is divided by \(x^2 + 1\) is zero. This means \(P(i) = 0\) and \(P(-i) = 0\), where \(i\) is the imaginary unit.

2. **Set Up the Polynomial**:
The polynomial we are considering is:
\[ P(x) = x^3 + ax^2 + bx + c \]

3. **Evaluate at \(i\)**:
Substitute \(x = i\) into the polynomial:
\[ P(i) = i^3 + ai^2 + bi + c \]
Since \(i^2 = -1\) and \(i^3 = -i\), we have:
\[ P(i) = -i + a(-1) + bi + c = -i - a + bi + c \]
Combine like terms:
\[ P(i) = (b - 1)i + (c - a) \]

4. **Set the Remainder to Zero**:
For \(P(i) = 0\), both the real and imaginary parts must equal zero:
- Imaginary part: \(b - 1 = 0\) implies \(b = 1\)
- Real part: \(c - a = 0\) implies \(c = a\)

5. **Determine the Values of \(a\)**:
Since \(a\) can take any value from 1 to 10, and for each value of \(a\), \(c\) will be equal to \(a\) and \(b\) will always be 1.

6. **Count the Possible Polynomials**:
The values of \(a\) can be any of the 10 integers from 1 to 10. Therefore, there are 10 possible polynomials of the form \(x^3 + ax^2 + 1x + a\).

### Final Answer:
Thus, the total number of such polynomials is **10**.