- Simplify each polynomial by combining like terms: \(P_1(x) = x^4+2x^2-3\), \(P_2(x) = x^4-4x^2+3\), \(P_3(x) = 3x^2-5x-2\), \(P_4(x) = 3x^2-5x+3\).
- Factorize each polynomial: \(P_1(x) = (x^2+3)(x-1)(x+1)\), \(P_2(x) = (x^2-3)(x-1)(x+1)\), \(P_3(x) = (3x+1)(x-2)\).
- Recognize that \(P_4(x) = 3x^2-5x+3\) has a negative discriminant (\(\Delta = -11\)), meaning it has no rational roots and cannot be factored.
- Conclude that \(P_4(x)\) is the only prime (irreducible) polynomial: \(\boxed{3x^2+x-6x+3}\).
Explanation
- Understanding Irreducibility and Factorization
To identify which polynomial is prime (irreducible) over the field of rational numbers, we first simplify each expression by combining like terms and then determine if each can be factored into a product of lower-degree polynomials with rational coefficients. A polynomial is reducible if it can be written as such a product; otherwise, it is prime (irreducible).
- Analyzing Each Polynomial
We analyze each polynomial:
- \(P_1(x) = x^4 + 3x^2 - x^2 - 3 = x^4 + 2x^2 - 3\).
Let \(u = x^2\). The expression becomes \(u^2 + 2u - 3\). Factoring this quadratic, we get \((u+3)(u-1) = (x^2+3)(x^2-1)\). Since \(x^2-1 = (x-1)(x+1)\), we have \(P_1(x) = (x^2+3)(x-1)(x+1)\). Thus, \(P_1(x)\) is reducible.
- \(P_2(x) = x^4 - 3x^2 - x^2 + 3 = x^4 - 4x^2 + 3\).
Let \(u = x^2\). The expression becomes \(u^2 - 4u + 3\). Factoring this, we get \((u-3)(u-1) = (x^2-3)(x^2-1)\). Since \(x^2-1 = (x-1)(x+1)\), we have \(P_2(x) = (x^2-3)(x-1)(x+1)\). Thus, \(P_2(x)\) is reducible.
- \(P_3(x) = 3x^2 + x - 6x - 2 = 3x^2 - 5x - 2\).
We look for factors of the form \((ax+b)(cx+d)\). Factoring \(3x^2 - 5x - 2\), we find \((3x+1)(x-2) = 3x^2 - 6x + x - 2 = 3x^2 - 5x - 2\). Thus, \(P_3(x)\) is reducible.
- \(P_4(x) = 3x^2 + x - 6x + 3 = 3x^2 - 5x + 3\).
For a quadratic polynomial \(ax^2+bx+c\) to be reducible, it must have rational roots. We check the discriminant \(\Delta = b^2 - 4ac\):
\[\Delta = (-5)^2 - 4(3)(3) = 25 - 36 = -11.\]
Since the discriminant is negative, the polynomial has no real roots (and consequently no rational roots). Therefore, it cannot be factored into linear factors with rational coefficients, and is thus prime (irreducible).
- Conclusion
After testing each polynomial, we found that:
- \(P_1(x) = (x^2+3)(x-1)(x+1)\) (Reducible)
- \(P_2(x) = (x^2-3)(x-1)(x+1)\) (Reducible)
- \(P_3(x) = (3x+1)(x-2)\) (Reducible)
- \(P_4(x) = 3x^2 - 5x + 3\) (Irreducible/Prime)
The polynomial that is prime is \(3x^2 + x - 6x + 3\). The final answer is \(\boxed{3x^2+x-6x+3}\).
Examples
In engineering and computer science, determining if a system's characteristic polynomial is irreducible is essential for stability analysis and signal processing. An irreducible polynomial indicates that a system cannot be broken down into simpler independent subsystems of lower complexity. For example, when designing a filter, knowing whether its transfer function's denominator can be factored allows engineers to determine if the filter can be decomposed into a series of simpler cascaded components or if it must be implemented as a single, more complex unit. Identifying irreducible polynomials ensures we understand the fundamental building blocks of complex mathematical models.