WHAT IS P(2) GIVEN OTHER VALUES OF THE CUBIC POLYNOMIAL FUNCTON? THE BEST METHOD.

WE USED THE GENERAL FORMULAR FOT CUBIC FUNCTION TO FIND THE VALUE OF P(2).
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[@alexandrezeddam7817]

Hi, unfortunately something is wrong in this video. While the solution you gave works, turns out any real solution works too! This is due to the fact that there are an infinite amount of degree 3 polynomial functions passing through 3 given points. Not onoy that, but you can specify any real value r and fixate f(2)=r and this will always give an unique solution. So where is the problem with your proof? Well, it happens at the veryyy start. You assume P(0)=0. Why? It's completely false! As an example, Q(X)=X^3+1 is a degree three polynomial where Q(0)=/=0. By saying that, you are implicitly considering a fourth equation in your equation set, being d=0 (you are falsely assuming P(0)=0). Hope this helps!

[@Center1ForMathematics]

Thanks for your concern. if you have any other solution other this one , please, send it to [e-mail] for review. Let me remind you one thing here, if you look at the general formular for cubic function or equation, you will understand that you will need equivalent number of equations as there are number of terms in the formula. Otherwise, to find the required P(2) will be based on assumption as you try to prove. Based on this formular for cubic function, when the function, itself is zero, the constant term will be zero. This simple concept formed the bases of solving this question based on only given three details. This is where I rest my case. We wait to hear from you soon.