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July 19th, 2017, 11:00 AM   #1
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p(x) for real x when p(p(x))+p(x)=x^4+3x^2+3

What I know:
deg of p(x) is 2 due to that degree of deg g(x)+p(x)=highest deg of the equation. Hence p(x) can be written as ax^2+bx+c.
But how do I continue?
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July 19th, 2017, 11:37 AM   #2
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If $p(x) = ax^2 + bx + c$, then what is $p(p(x))$?
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July 19th, 2017, 12:13 PM   #3
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p(p(x)) is then p(ax^2+bx+c) of course.
Then we have p(ax^2+bx+c) + p(x) = x^4+3x^2+3. I have tried to go on with finding p(x), but I have three unknowns...
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July 19th, 2017, 12:17 PM   #4
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So can you simplify $p(ax^2 + bx + c)$??
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July 19th, 2017, 12:20 PM   #5
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Yes, into $\displaystyle p((x+m)(ax+n))+p(x)$. But do not understand how that is helping me?

Last edited by oscar3; July 19th, 2017 at 12:37 PM.
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July 19th, 2017, 12:23 PM   #6
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Why not simply say that
$$p(p(x)) = a(p(x))^2 + bp(x) + c = a(ax^2 + bx + c)^2 + b(ax^2 + bx + c) + c$$
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July 19th, 2017, 12:35 PM   #7
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Smart!

So now let's simplify:
$\displaystyle a(ax^2+bx+c)^2+b(ax^2+bx+c)+c+(ax^2+bx+c)=a(ax^2+b x+c)^2+(b+1)(ax^2+bx+c)+c$
Or am I wrong?

Last edited by skipjack; July 19th, 2017 at 09:45 PM.
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July 19th, 2017, 12:37 PM   #8
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Quote:
Originally Posted by oscar3 View Post
Smart!

So now let's simplify:
$\displaystyle a(ax^2+bx+c)^2+b(ax^2+bx+c)+c+(ax^2+bx+c)=a(ax^2+b x+c)^2+(b+1)(ax^2+bx+c)+c$
Or am I wrong?
Right; this is $p(p(x)) + p(x)$.
Thanks from oscar3

Last edited by skipjack; July 19th, 2017 at 10:02 PM.
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July 19th, 2017, 12:38 PM   #9
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But, should I not know the coefficients of it?

Last edited by skipjack; July 19th, 2017 at 09:35 PM.
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July 19th, 2017, 12:41 PM   #10
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Never mind, it is not necessary
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