Solve (with respect to the only variable, x) the equation
  (2x-1)·(2x+1)·(3x+1) = 0

(2x-1)(2x+1)(3x+1) = 0 when at least one of 2x-1, 2x+1 e 3x+1 is 0. So there are three possibilities for the equation to be true:
2x-1 = 0, i.e. 2x = 1, i.e. x = 1/2;
2x+1 = 0, i.e. 2x = −1, i.e. x = −1/2;
3x+1 = 0, i.e. 3x = −1, i.e. x = −1/3.
  These are the three solutions.

    The question seems very easy, however in an entrance test administered to a large sample (7300 people) of the freshmen of the scientific faculties of the Universities of Genoa and Pisa (in 1982 and 1983) there were only 64% correct answers , and not always obtained in the simple way illustrated above.
    Many students multiplied the three polynomials obtaining a 3rd degree polynomial equation which they then tried to solve with various decomposition techniques or in front of which they claimed not to know the solution formula of the 3rd degree equations!
    There is the bad habit (perhaps cultivated by teaching) to face everything with manipulations or trying to remember some "recipe", instead of trying to analyze the proposed problems in a more reflective way (and, in this case, think about the meaning of "solution of an equation").

    Similarly, in front of the study of the curve y = 3(x−2)² + 4  it is counterproductive to expand the polynomial 3(x−2)² + 4, but many pupils proceed in this way, in order to use their baggage of formulas for the 2nd degree equations, as they have been used to.