document.write( "Question 1207386: Suppose P(x) is a quadratic polynomial (highest degree of 2) satisfying
\n" ); document.write( "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.\r
\n" ); document.write( "\n" ); document.write( "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!
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Algebra.Com's Answer #845239 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "Let \"P%28x%29=ax%5E2%2Bbx%2Bc\"

\n" ); document.write( "Then

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\n" ); document.write( "\"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\"

\n" ); document.write( "We need to have

\n" ); document.write( "\"P%28P%28x%29%29-P%28x%29%5E2=x%5E2%2Bx%2B2016\"

\n" ); document.write( "\"%28a-1%29%28ax%5E2%2Bbx%2Bc%29%5E2%2Bb%28ax%5E2%2Bbx%2Bc%29%2Bc=x%5E2%2Bx%2B2016\"

\n" ); document.write( "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.

\n" ); document.write( "Then

\n" ); document.write( "\"b%28ax%5E2%2Bbx%2Bc%29%2Bc=x%5E2%2Bx%2B2016\"

\n" ); document.write( "Equating the coefficients of x^2 on the two sides of the equation, we need to have

\n" ); document.write( "\"ab=1\"

\n" ); document.write( "and since a=1, b=1 also. And now, with a=1 and b=1, we have

\n" ); document.write( "\"x%5E2%2Bx%2Bc%2Bc=x%5E2%2Bx%2B2016\"
\n" ); document.write( "\"2c=2016\"
\n" ); document.write( "\"c=1008\"

\n" ); document.write( "And we have the polynomial we need.

\n" ); document.write( "ANSWER: \"P%28x%29=x%5E2%2Bx%2B1008\"

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