How many real solutions has the following equation with respect to X?
        (X6 + π)·(5X + 1)·(X + 2)2·(X3 – 1) = 0
(A) it is an equation of degree 12: a solution formula is not known; therefore the question cannot be answered
(B) 3 		(C) 12
(D) 4
(E) 6

(X6 + π)·(5X + 1)·(X + 2)2·(X3 – 1) = 0  when  5X+1=0  or  X+2=0  o  X3-1=0 (X6+π never is 0). So -1/5, −2 and 1 are the only (real) numbers which, substituted for x, make the equation true. The correct answer is "3 solutions".

The solutions can also be found with WolframAlpha by introducing:
solve (x^6+pi)*(5*x+1)*(x+2)^2*(x^3-1) = 0 for x real

Note. It is not true that every polynomial eq. of degree N has N solutions!!!  x100=0 è is true only for x=0; in some particular mathematical areas, to address some questions (eg specify how close to the solution the curve y=… "squashes" on the x axis), it is useful to introduce the concept of multiplicity, and to say that, e.g., for this equation, 0 is a solution of multiplicity 100  (we repeat: it would be an enormous nonsense, especially in teaching, to say that it has 100 coincident solutions:  100 ≠ 1!).
Note also that the formulation of the answer A is incorrect:  the Ruffini-Abel theorem does not state that for the polynomial equations in x P(x)=0 of degree greater than 4 there are no formulas or other resolutive procedures, but that there are no general formulas that allow solutions (with respect to x) to be expressed as functions of the coefficients of P(x) obtained by composing only the four root operations and (square root or higher order) extractions!

In a test submitted to about forty graduates in scientific faculties (in 2000) the question had only 38% of correct answers.  19% of graduates chose "12 solutions", 16% chose "4 solutions". 27% preferred not to respond.