Which polynomials are prime? Check all of the boxes that apply.
[ ] $x^2+9$
[ ] $x^2-9$
[ ] $x^2+3 x+9$
[ ] $-2 x^2+8$
A polynomial \(P(x)\) with real coefficients is considered prime, or irreducible, over the set of real numbers if it cannot be expressed as the product of two non-constant polynomials with real coefficients. For a quadratic polynomial of the form \(ax^2 + bx + c\), the irreducibility is determined by its discriminant \(\Delta = b^2 - 4ac\). If \(\Delta < 0\), the quadratic has no real roots and cannot be factored into real linear factors, making it irreducible. If \(\Delta \geq 0\), the polynomial has real roots and can be factored into at least two linear factors. Constant factors, such as \(-2\) in \(-2x^2+8\), do not affect the irreducibility of the polynomial.
The polynomial \(P_1(x) = x^2 + 9\) has coefficients \(a = 1, b = 0,\) and \(c = 9\). Calculating the discriminant:
\[\Delta = b^2 - 4ac = 0^2 - 4(1)(9) = -36.\]
Since \(\Delta = -36 < 0\), the polynomial has no real roots. Therefore, \(x^2 + 9\) cannot be factored into real linear factors, meaning it is a prime polynomial.
The polynomial \(P_2(x) = x^2 - 9\) has coefficients \(a = 1, b = 0,\) and \(c = -9\). Calculating the discriminant:
\[\Delta = b^2 - 4ac = 0^2 - 4(1)(-9) = 36.\]
Since \(\Delta = 36 > 0\), the polynomial has two distinct real roots. Using the difference of squares identity, we can factor it:
\[x^2 - 9 = (x-3)(x+3).\]
Because it is a product of two non-constant polynomials, \(x^2 - 9\) is not prime.
The polynomial \(P_3(x) = x^2 + 3x + 9\) has coefficients \(a = 1, b = 3,\) and \(c = 9\). Calculating the discriminant:
\[\Delta = b^2 - 4ac = 3^2 - 4(1)(9) = 9 - 36 = -27.\]
Since \(\Delta = -27 < 0\), the polynomial has no real roots. Therefore, \(x^2 + 3x + 9\) cannot be factored into non-constant real polynomials, making it a prime polynomial.
The polynomial \(P_4(x) = -2x^2 + 8\) can be simplified by factoring out the constant \(-2\):
\[-2x^2 + 8 = -2(x^2 - 4).\]
The expression \(x^2 - 4\) is a difference of squares, which factors further:
\[-2(x^2 - 4) = -2(x-2)(x+2).\]
Since the polynomial can be expressed as the product of non-constant polynomials (e.g., \((x-2)\) and \((-2x-4)\)), it is not prime.
Based on the analysis, the prime (irreducible) polynomials are those with a negative discriminant, which ensures they cannot be factored into real non-constant factors. We found that \(x^2+9\) and \(x^2+3x+9\) are irreducible.
The prime polynomials are:
\(\boxed{x^2+9, x^2+3x+9}\)
In engineering and physics, checking the irreducibility of polynomials is essential when solving differential equations. For instance, in the study of vibrational systems or control theory, the roots of the "characteristic polynomial" determine the stability of the system. An irreducible quadratic polynomial corresponds to complex roots, which represent oscillating behavior, whereas a reducible quadratic indicates real roots, which might lead to exponential decay or growth. Identifying prime polynomials helps scientists and engineers categorize the fundamental behaviors of physical systems quickly without performing full factorization every time.