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

Step 1: Understanding the Concept:
For a quadratic to divide a cubic, the cubic must be of the form \( (x^2 + 2)(x + k) \). We equate coefficients to find relationships between \(b, c,\) and \(d\).
Step 2: Key Formula or Approach:
Let \(x^3 + bx^2 + cx + d = (x^2 + 2)(x + k)\) for some integer \(k\).
Step 3: Detailed Explanation:
Expand the right side:
\[ (x^2 + 2)(x + k) = x^3 + kx^2 + 2x + 2k \] Comparing coefficients with \(x^3 + bx^2 + cx + d\):
1. \(b = k\)
2. \(c = 2\)
3. \(d = 2k = 2b\)
Constraints: \(b, c, d \in \{1, 2, 3, \dots, 20\}\).
From \(c=2\), this is always a natural number \(\le 20\).
From \(d = 2b\), we need \(2b \le 20 \implies b \le 10\).
Since \(b \in N\), \(b\) can be any value in \(\{1, 2, \dots, 10\}\).
For each value of \(b\), \(d\) is uniquely determined as \(2b\), and \(c\) is fixed at 2.
There are 10 such possible triplets.
Step 4: Final Answer:
The number of such triplets is 10.