Hi, andrewkirk. Thanks for commenting. I see your point. However, I did not understand your last statement. Correct me, if I am not mistaken, but by substituting ##y=-1/2## into ##P(x,y)## we will have ##P(x,-1/2) \equiv \tilde{P}(x)=50x^2+100x+50##, or, simply,##\tilde{P}(x)= 50(x+1)^2##. Based...