Question

The polynomial \(x^2+9\) is reducible over ____.

• A. \(\mathbb{N}\)
• B. \(\mathbb{W}\)
• C. \(\mathbb{R}\)
• D. \(\mathbb{Z}\)

Hint:

A quadratic \(x^2+a^2\) is irreducible over \(\mathbb{R}\) and \(\mathbb{Z}\), but reducible over \(\mathbb{C}\) as \((x+ai)(x-ai)\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
A polynomial is said to be reducible over a set if it can be expressed as the product of two non-constant polynomials whose coefficients belong to that set. Otherwise, it is called irreducible. 

Step 2: Key Formula or Approach
Consider the polynomial \[ x^2+9. \] To determine whether it is reducible, we check if it can be factored using coefficients from the given number system. 

Step 3: Detailed Explanation
Over the real numbers, \[ x^2+9=0 \] gives \[ x^2=-9, \] which has no real solution. Hence, \[ x^2+9 \] cannot be factorized over \(\mathbb{R}\). Therefore, it is also irreducible over the smaller sets \(\mathbb{N}\), \(\mathbb{W}\), and \(\mathbb{Z}\). However, over the integers, \[ x^2+9=x^2+0x+9. \] Since \[ 9=1\times9=3\times3, \] there are no integers \(a\) and \(b\) such that \[ (a+b)=0 \quad\text{and}\quad ab=9. \] Thus, \[ x^2+9 \] cannot be written as \[ (x+a)(x+b), \] where \(a,b\in\mathbb{Z}\). Therefore, the polynomial is irreducible over \(\mathbb{Z}\), \(\mathbb{R}\), \(\mathbb{N}\), and \(\mathbb{W}\). Hence, the answer key stating Option (4) is incorrect. The polynomial is not reducible over any of the given sets. It becomes reducible only over the complex numbers: \[ x^2+9=(x+3i)(x-3i). \] 

Step 4: Conclusion
The polynomial is reducible over \(\mathbb{C}\), but not over any of the given options. Therefore, among the given choices, there is no correct option. If the official key marks Option (4), the key is incorrect. 

Final Answer: The given answer key is incorrect. The polynomial \(x^2+9\) is irreducible over \(\mathbb{N}\), \(\mathbb{W}\), \(\mathbb{Z}\), and \(\mathbb{R}\); it is reducible only over \(\mathbb{C}\).