Answer:
In general, determining whether a polynomial is prime or not is a complex task, especially for polynomials of higher degree. However, we can use some techniques to eliminate some of the options provided.
The first and second polynomials both have a common factor of (x+1), so they are not prime. We can factor them as follows:
Ox3³+3x²+2x+6 = (x+1)(x^2+2x+6)
x³ + 3x²-2x-6 = (x+1)(x^2+2x-6)
The third and fourth polynomials are both quadratic, but they cannot be factored easily using integer coefficients. However, we can apply the quadratic formula to check if they have any real roots. If they do, then they are not prime.
For the third polynomial, we have:
10x² - 4x + 3x + 6 = 10x² - x + 6
Using the quadratic formula, we have:
x = [1 ± sqrt(1 - 4*10*6)] / (2*10)
x = (-1 ± sqrt(-239)) / 20
Since the discriminant is negative, the roots are imaginary and the polynomial has no real roots. Therefore, it is a candidate for a prime polynomial.
For the fourth polynomial, we have:
10x² + 6x - 6
Using the quadratic formula, we have:
x = [-6 ± sqrt(6^2 - 4*10*(-6))] / (2*10)
x = (-3 ± sqrt(39)) / 5
Since the roots are real, the polynomial is not prime.
Therefore, the only candidate for a prime polynomial among the given options is:
10x² - x + 6