SOLUTION: Suppose P(x) is a quadratic polynomial (highest degree of 2) satisfying P(P(x)) - (P(x))^2 = x^2 + x + 2016 for all real x. Find P(x). Express the answer as a quadratic polynomial

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Suppose P(x) is a quadratic polynomial (highest degree of 2) satisfying P(P(x)) - (P(x))^2 = x^2 + x + 2016 for all real x. Find P(x). Express the answer as a quadratic polynomial      Log On


   

Question 1207386: Suppose P(x) is a quadratic polynomial (highest degree of 2) satisfying
P(P(x)) - (P(x))^2 = x^2 + x + 2016 for all real x. Find P(x). Express the answer as a quadratic polynomial with highest degree of 2.
I've been stuck on this one for a while now and haven't found anything useful to solve. Can anyone help me out? Thanks in advance!
Answer by greenestamps(13367) About Me  (Show Source):
You can put this solution on YOUR website!


Let P%28x%29=ax%5E2%2Bbx%2Bc

Then



P%28P%28x%29%29-P%28x%29%5E2=%28a-1%29%28ax%5E2%2Bbx%2Bc%29%5E2%2Bb%28ax%5E2%2Bbx%2Bc%29%2Bc

We need to have

P%28P%28x%29%29-P%28x%29%5E2=x%5E2%2Bx%2B2016

%28a-1%29%28ax%5E2%2Bbx%2Bc%29%5E2%2Bb%28ax%5E2%2Bbx%2Bc%29%2Bc=x%5E2%2Bx%2B2016

On the left hand side of the equation, %28ax%5E2%2Bbx%2Bc%29%5E2 is going to give x%5E4 and x%5E3 terms; but there are no terms of that degree on the right hand side. That means we must have (a-1)=0, or a=1.

Then

b%28ax%5E2%2Bbx%2Bc%29%2Bc=x%5E2%2Bx%2B2016

Equating the coefficients of x^2 on the two sides of the equation, we need to have

ab=1

and since a=1, b=1 also. And now, with a=1 and b=1, we have

x%5E2%2Bx%2Bc%2Bc=x%5E2%2Bx%2B2016
2c=2016
c=1008

And we have the polynomial we need.

ANSWER: P%28x%29=x%5E2%2Bx%2B1008