Question:medium

Total number of polynomials of the form \(x^3 + ax^2 + bx + c\) that are divisible by \(x^2 + 1\), where \(a,b,c \in \{1, 2, 3, \dots, 10\}\) is equal to

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Polynomial division condition leads to relationships between coefficients.
Updated On: May 24, 2026
  • 90
  • 45
  • 5
  • 10
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The Correct Option is D

Solution and Explanation

We need to determine the number of polynomials of the form \(x^3 + ax^2 + bx + c\) that are divisible by \(x^2 + 1\), where \(a, b, c \in \{1, 2, 3, \dots, 10\}\).

Since \(x^2 + 1\) is a factor of the polynomial, it must divide it completely, leading to the condition:

1. The remainder when this polynomial is divided by \(x^2 + 1\) should be zero.

2. The roots of the polynomial \(x^2 + 1 = 0\) are \(x = i\) and \(x = -i\).

This means both \(i\) and \(-i\) should satisfy the polynomial equation:

\(x^3 + ax^2 + bx + c = 0\) for \(x = i\) and \(x = -i\).

Substitute \(x = i\) into the polynomial:

\(i^3 + ai^2 + bi + c = 0\)

We know \(i^2 = -1\) and \(i^3 = -i\), so the equation becomes:

\(-i - a + bi + c = 0\)

- which is equivalent to \((b - 1)i + (c - a) = 0\).

This implies that:

\(b - 1 = 0 \quad \text{and} \quad c - a = 0\)

We conclude:

  • \(b = 1\)
  • \(c = a\)

Now, substituting \(x = -i\) into the polynomial:

\((-i)^3 + a(-i)^2 + b(-i) + c = 0\)

Simplify this using \((-i)^2 = -1\) and \((-i)^3 = i\):

\(i + a(-1) - bi + c = 0\)

- which becomes \((1 - b)i + (c - a) = 0\)

We notice it matches our previous conditions:

  • \(b = 1\)
  • \(c = a\)

Given that \(a, b, c \in \{1, 2, 3, \dots, 10\}\), let's list the valid polynomials:

Since \(b = 1\), and \(c = a\), we can choose \(a\) in 10 different ways (from 1 to 10).

Therefore, the total number of such polynomials that satisfy these conditions is 10.

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