To determine which expression is a prime polynomial, we need to understand what makes a polynomial "prime". A prime polynomial over the integers is one that cannot be factored into the product of two non-constant polynomials with integer coefficients.
Let's analyze each of the given options step-by-step to see if they can be factored:
### Option A: [tex]\(x^3 - 1\)[/tex]
The polynomial [tex]\(x^3 - 1\)[/tex] can be factored using the difference of cubes formula:
[tex]\[ x^3 - 1 = (x - 1)(x^2 + x + 1) \][/tex]
So, [tex]\(x^3 - 1\)[/tex] is not a prime polynomial because it can be factored.
### Option B: [tex]\(x^2 + 1\)[/tex]
To test if [tex]\(x^2 + 1\)[/tex] can be factored over the integers, let's see if there are any integer solutions to the equation:
[tex]\[ x^2 + 1 = 0 \implies x^2 = -1 \][/tex]
Since there are no real roots to this equation (as the square of a real number cannot be negative), [tex]\(x^2 + 1\)[/tex] cannot be factored further over the integers. Hence, [tex]\(x^2 + 1\)[/tex] is a prime polynomial.
### Option C: [tex]\(x^3 + 1\)[/tex]
The polynomial [tex]\(x^3 + 1\)[/tex] can be factored using the sum of cubes formula:
[tex]\[ x^3 + 1 = (x + 1)(x^2 - x + 1) \][/tex]
So, [tex]\(x^3 + 1\)[/tex] is not a prime polynomial because it can be factored.
### Option D: [tex]\(x^2 - 1\)[/tex]
The polynomial [tex]\(x^2 - 1\)[/tex] can be factored using the difference of squares formula:
[tex]\[ x^2 - 1 = (x - 1)(x + 1) \][/tex]
So, [tex]\(x^2 - 1\)[/tex] is not a prime polynomial because it can be factored.
Based on the analysis above, the only prime polynomial among the given options is:
B. [tex]\(x^2 + 1\)[/tex]
