Step 1: Understanding the Concept:
A polynomial is reducible over a set (or ring) if it can be factored into two non-constant polynomials with coefficients from that set. A polynomial that cannot be factored in this way is called irreducible. We need to determine over which of the given sets of numbers the polynomial \(x^2+9\) can be factored.
Step 2: Key Formula or Approach:
To factor \(x^2+9\), we would need to find its roots. The roots of \(x^2+9=0\) are \(x^2=-9\), which gives \(x = \pm \sqrt{-9} = \pm 3i\).
The factorization is \((x - 3i)(x + 3i)\).
A polynomial is reducible over a set if the coefficients of its factors belong to that set.
Let's check each option:
- \(\mathbb{N}\) (Natural Numbers), \(\mathbb{W}\) (Whole Numbers), \(\mathbb{Z}\) (Integers), \(\mathbb{R}\) (Real Numbers): These sets do not contain the imaginary numbers \(3i\) and \(-3i\).
- A polynomial of degree 2 or 3 is reducible over a field F if and only if it has a root in F.
Step 3: Detailed Explanation:
The polynomial is \(p(x) = x^2+9\).
Reducibility over \(\mathbb{R}\) (Real Numbers):
For \(x^2+9\) to be reducible over \(\mathbb{R}\), it must have real roots. The equation \(x^2+9=0\) has no real solutions, since \(x^2 = -9\) is impossible for any real number \(x\). The discriminant is \(\Delta = 0^2 - 4(1)(9) = -36 < 0\), confirming no real roots. Therefore, \(x^2+9\) is irreducible over \(\mathbb{R}\). Statement (C) is incorrect.
Reducibility over \(\mathbb{Z}\) (Integers):
Since \(\mathbb{Z}\) is a subset of \(\mathbb{R}\), if the polynomial is irreducible over \(\mathbb{R}\), it must also be irreducible over \(\mathbb{Z}\). If it could be factored over \(\mathbb{Z}\) as \((ax+b)(cx+d)\) with integer coefficients, then it could also be factored over \(\mathbb{R}\), which we know is not possible. Therefore, \(x^2+9\) is irreducible over \(\mathbb{Z}\).
Reducibility over \(\mathbb{N}\) and \(\mathbb{W}\):
These are not rings, so the concept of reducibility is not standardly defined. But if we interpret it as factoring into polynomials with coefficients in these sets, it's still irreducible.
Conclusion on Discrepancy:
None of the provided options appear to be correct under the standard definition of reducibility. The polynomial \(x^2+9\) is irreducible over \(\mathbb{N}, \mathbb{W}, \mathbb{Z},\) and \(\mathbb{R}\). It is only reducible over the complex numbers \(\mathbb{C}\), since \(x^2+9 = (x-3i)(x+3i)\).
The question or the options (or the provided answer key) are flawed. If there is a misunderstanding, perhaps "reducible over Z" is being used in a non-standard way. However, based on all standard definitions in algebra, this polynomial is irreducible over all the given sets.
Given the provided answer key marks Z, there is a profound error in the question. There is no standard mathematical context where \(x^2+9\) is reducible over the integers but not the reals.
Step 4: Final Answer:
The polynomial \(x^2+9\) is irreducible over \(\mathbb{R}\) and \(\mathbb{Z}\). None of the options are correct. The question is flawed.