Question

The total number of polynomials of the form \[ x^{3}+ax^{2}+bx+c \] which are divisible by \(x^{2}+1\), where \(a,b,c\in\{1,2,3,\dots,10\}\) is:

• A. 120
• B. 45
• C. 10
• D. 15

Hint:

You can also solve this using polynomial long division! Dividing $x^3+ax^2+bx+c$ by $x^2+1$ leaves a remainder of $(b-1)x + (c-a)$. For the polynomial to be perfectly divisible, this remainder must be zero, which instantly gives you the equations $b=1$ and $c=a$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
If a polynomial \( P(x) \) is divisible by \( x^2 + 1 \), then the roots of \( x^2 + 1 = 0 \) must also be roots of \( P(x) \). The roots are \( x = i \) and \( x = -i \).
Step 2: Key Formula or Approach:
1. Set \( P(i) = 0 \).
2. Separate the resulting equation into real and imaginary parts.
3. Solve for conditions on \( a, b, c \).
Step 3: Detailed Explanation:
Let \( P(x) = x^3 + ax^2 + bx + c \).
If \( x^2 + 1 \) divides \( P(x) \), then \( P(i) = 0 \).
\[ (i)^3 + a(i)^2 + b(i) + c = 0 \]
Since \( i^2 = -1 \) and \( i^3 = -i \):
\[ -i - a + bi + c = 0 \]
Rearrange into real and imaginary components:
\[ (c - a) + i(b - 1) = 0 \]
For this to be zero, both the real and imaginary parts must be zero:
1. \( b - 1 = 0 \implies b = 1 \).
2. \( c - a = 0 \implies c = a \).
Now we look at the constraints: \( a, b, c \in \{1, 2, \dots, 10\} \).
- For \( b \), there is only 1 choice: \( b = 1 \).
- For \( a \) and \( c \), they must be equal. We can choose any value from 1 to 10 for \( a \), and \( c \) will automatically be the same value. There are 10 choices for the pair \( (a, c) \).
Total polynomials \( = 1 \times 10 = 10 \).
Step 4: Final Answer:
There are 10 such polynomials.