Final answer:
Upon examining the given polynomials, B) x² - 9x + 3 and E) 2 - 3r are identified to be prime polynomials as they cannot be factored further. Other options either can be factored or are not proper polynomials as given.
Step-by-step explanation:
A prime polynomial is a polynomial that cannot be factored into the product of two non-constant polynomials. Identifying prime polynomials among the options given requires checking each one to see if it can be factored.
A) 4 + 3y²r - 1 - This is not a polynomial in standard form. To consider it as a prime polynomial, it would need to be re-written, but as it stands and assuming a typo, it cannot be considered without further clarification.
B) x² - 9x + 3 - This is a quadratic polynomial. To check if it's prime, you would look for two numbers that multiply to give the product of the coefficient of x² (which is 1) and the constant term (which is 3), and add up to the coefficient of x (which is -9). There are no such integers, making this polynomial prime.
C) 13 - 512y³ - This polynomial is a difference of two cubes, as 13 is not a perfect cube and 512 is 8 cubed. Therefore, it can be factored as (2 - 8y)(4 + 16y + 64y²), thus it is not prime.
D) 9y - 4y² + 12y³ - This is a cubic polynomial. To determine if it's prime, try to factor by grouping or using other factoring methods. It appears that factoring out a common factor of y, we get y(9 - 4y + 12y²). However, the remaining trinomial can be further factored, so this polynomial is not prime.
E) 12 - 3r - 10 - This can be further simplified to 2 - 3r, which cannot be factored any further into a product of polynomials with integer coefficients. Therefore, it is a prime polynomial.
In conclusion, the prime polynomials from the given options are B) x² - 9x + 3 and E) 2 - 3r.