Let p be a prime number. The quadratic equation having its roots as factors of p is(a) x2 –px +p=0 (b) x2–(p+1)x +p=0 (c) x2+(p+1)x +p=0 (d) x2 –px+p+1=0

The roots of a quadratic equation are given by the formula x = [-b ± sqrt(b^2 - 4ac)] / 2a.

For a prime number p, its only factors are 1 and p itself.

Let's examine each of the given equations:

(a) x^2 –px +p=0 Here, a=1, b=-p, c=p. The roots are given by x = [p ± sqrt((-p)^2 - 41p)] / 2*1 = [p ± sqrt(p^2 - 4p)] / 2. These roots are not necessarily factors of p, so this equation is not correct.

(b) x^2–(p+1)x +p=0 Here, a=1, b=-(p+1), c=p. The roots are given by x = [(p+1) ± sqrt((-(p+1))^2 - 41p)] / 2*1 = [(p+1) ± sqrt((p^2 + 2p + 1) - 4p)] / 2 = [(p+1) ± sqrt(p^2 - 2p + 1)] / 2. These roots are not necessarily factors of p, so this equation is not correct.

(c) x^2+(p+1)x +p=0 Here, a=1, b=p+1, c=p. The roots are given by x = [-(p+1) ± sqrt((p+1)^2 - 41p)] / 2*1 = [-(p+1) ± sqrt((p^2 + 2p + 1) - 4p)] / 2 = [-(p+1) ± sqrt(p^2 - 2p + 1)] / 2. These roots are not necessarily factors of p, so this equation is not correct.

(d) x^2 –px+p+1=0 Here, a=1, b=-p, c=p+1. The roots are given by x = [p ± sqrt((-p)^2 - 41(p+1))] / 2*1 = [p ± sqrt(p^2 - 4p - 4)] / 2. These roots are not necessarily factors of p, so this equation is not correct.

Therefore, none of the given equations have their roots as factors of a prime number p.