Question

Find possible no. of triplets (b, c, d), such that \(x^2 + 2\) is divisor of \(x^3 + bx^2 + cx + d\) & b, c, d \( \le \) 20 & b, c, d \( \in \) N :

Hint:

If a quadratic without an \(x\) term (like \(x^2+2\)) divides a cubic, the linear term of the cubic must be exactly twice the constant term of the linear factor. This allows for quick constraint checking.

Answer 1

Correct Answer: 10

To find the number of triplets \((b, c, d)\) such that \(x^2 + 2\) is a divisor of \(x^3 + bx^2 + cx + d\), we apply the properties of polynomial division. Specifically, for \(x^2 + 2\) to divide \(x^3 + bx^2 + cx + d\), the remainder when dividing by \(x^2 + 2\) must be zero. This implies that:

1. First, express \(x^3 + bx^2 + cx + d\) as \( (x^2 + 2) \cdot Q(x) + R(x)\), where \(R(x) = 0\) since the remainder must be zero.
2. Since \(R(x)\) is zero and typically of degree less than \(x^2\), it simplifies to the condition:

Substitute \(x^2 = -2\) in the expression \(x^3 + bx^2 + cx + d\):

\(x^3 = x \cdot x^2 = x(-2) = -2x\).

Thus, \(x^3 + bx^2 + cx + d = -2x + b(-2) + cx + d = (-2x - 2b + cx + d)\).

For the polynomial to be zero for \(x^2 + 2\), the coefficients of \(x\) and the constant term must be zero:

1. \(-2 + c = 0 \rightarrow c = 2\). 
2. \(-2b + d = 0 \rightarrow d = 2b\).

Since \(b, c, d \in \mathbb{N}\) and \(b, c, d \leq 20\), we determine feasible values:

1. \(c = 2\) (fixed by condition 1).
2. \(d = 2b\) must satisfy \(d \leq 20\), so \(b \leq 10\).

The possible values of \(b\) are integers from 1 to 10. 
Thus, there are 10 possible values for \(b\), each corresponding to a distinct triplet \((b, 2, 2b)\).

Total number of triplets: 10

The calculated range is within the expectation of the range (10,10).