SOLUTION: IF X AND P ARE ZEROES OF THE POLYNOMIAL, F(x)=x2_P(x+1)-c THEN FIND OUT (x+1)(P+1).. sir i have asked this question but there was no reply pls sir i need ur help to solv

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: IF X AND P ARE ZEROES OF THE POLYNOMIAL, F(x)=x2_P(x+1)-c THEN FIND OUT (x+1)(P+1).. sir i have asked this question but there was no reply pls sir i need ur help to solv      Log On


   



Question 202131: IF X AND P ARE ZEROES OF THE POLYNOMIAL,
F(x)=x2_P(x+1)-c
THEN FIND OUT (x+1)(P+1)..

sir i have asked this question but there was no reply pls sir i need ur help to solve this question i'll be thankful


Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Perhaps the reason that there has been no reply is that much of what you have provided makes no sense.

IF X AND P ARE ZEROES OF THE POLYNOMIAL,
It is very unusual (and confusing) to say "X" if a zero of a polynomial and then use "x" to define the polynomial. Usually case (upper or lower) is meaningless. In other words, "X" and "x" usually mean the same thing. But in your problem they appear to represent different things.

F(x)=x2_P(x+1)-c
What is the "_"? Is it supposed to be "+"?
What is "x2"? Is "x2" meant to be "x squared"? If yes, then write it this way: x^2

THEN FIND OUT (x+1)(P+1)..
What is the ".."?
Is the problem to find (x+1)*(P+1)? If so, then the rest of the problem means nothing. Just use FOIL to multiply this:
%28x+%2B+1%29%2A%28P+%2B+1%29+=+x%2AP+%2B+x%2A1+%2B+1%2AP+%2B+1%2A1+=+xP+%2B+x+%2B+P+%2B+1
Or are we supposed to find F(x+1) and F(P+1)? Or F((x+1)*(P+1))?

Provide a problem that can be understood and you will probably get a reply.