SOLUTION: Given that x^2 + px +q and 3x^2 + q have a common factor (x-b) where p,q and b are non zero.Show that 3p^2 + 4q = 0

From the condition, we can conclude that "b" is the root for each given polynomial of x,
due to the Remainder theorem.  So,  

    b^2 + pb + q = 0      (1)

and  

    3b^2 + q = 0.         (2)


From (2),  q = -3b^2.  Substitute it into (1)

    b^2 + pb - 3b^2 = 0,

or

    2b^2 = pb.


Since b is not equal to zero ( ! given ! ), we can divide both sides by b in the last equation. We get then

    2b = p.


It implies  

    (2b)^2 = p^2  -->  4b^2 = p^2  -->  12b^2 = 3p^2.    (3).


From the other side,  from (2) we have  

    3b^2 = -q    -->                    12b^2 = -4q.     (4)


In (3) and (4), left sides are identical, so their right sides are equal:  3p^2 = -4q.

which means  3p^2 + 4q = 0.    QED.