0. (a) Find all solutions of the quadratic equation x2 + 5x + 6 = 0 in Z12.[4 marks](b) Let f (x) = x3 + 9x2 + 1 ∈ Z11[x]. Find three distinct zeroes of f (x) andhence write f (x) as a product of three distinct linear factors.

(a) To find the solutions of the quadratic equation x^2 + 5x + 6 = 0 in Z12, we first need to find the roots of the equation.

The roots of a quadratic equation ax^2 + bx + c = 0 can be found using the quadratic formula x = [-b ± sqrt(b^2 - 4ac)] / 2a.

Here, a = 1, b = 5, and c = 6.

So, x = [-5 ± sqrt((5)^2 - 416)] / 2*1 = [-5 ± sqrt(25 - 24)] / 2 = [-5 ± sqrt(1)] / 2 = [-5 ± 1] / 2

So the solutions are x = -3 and x = -2.

However, we are looking for solutions in Z12, which means we are looking for solutions modulo 12.

So, x = -3 mod 12 = 9 and x = -2 mod 12 = 10.

So, the solutions of the equation x^2 + 5x + 6 = 0 in Z12 are x = 9 and x = 10.

(b) To find the zeroes of the function f(x) = x^3 + 9x^2 + 1 in Z11[x], we need to find the values of x for which f(x) = 0.

We can do this by trying all possible values of x in Z11, which are 0, 1, 2, ..., 10.

By trying these values, we find that the zeroes of f(x) are x = 0, x = 1, and x = 2.

Therefore, we can write f(x) as a product of three distinct linear factors:

f(x) = (x - 0)(x - 1)(x - 2) = x(x - 1)(x - 2).