# Solving a difficult polynomial remainder question..

"Let $P(x)$ be a polynomial such that

$$P(P(x))=16x-5$$

What is $P(x-3)$ when divided by $x-5$?"

I have never seen a polynomial inside a polynomial before in a question so I have no idea how to proceed. I tried to give it a shot and I ended up at here:

$P(P2)=27$

The answer must be $7$. I would appreciate if someone showed me the way to solve this polynomial.

• First step: find $P$. Do you know what $P(P(x))$ means? Nov 14 '17 at 18:52
• $P(P(ax+b))$ I'm assuming? Nov 14 '17 at 18:54
• No. $x$ and $ax+b$ are different things, so $P(P(x))$ and $P(P(ax+b))$ are, at least a priori, different things as well. Rather, it's $P(ax+b)$, which again becomes $a(ax+b) + b$ (assuming $P(x) = ax+b$). Nov 14 '17 at 18:55

• If $P$ has degree 0, you cannot get a linear result in $P(P(x))$. If $P$ has degree more than 1, $P(P(x))$ will have an even higher degree. So it must be that $P$ has degree 1, and hence, $P(x) = ax+b$ for some $a,b \in \mathbb{R}$.
• Can you plug in this definition and compute $P(P(x))$ and then derive 2 equations for $a,b$?
• I found that $a=4$ and $b=-1$ but I don't know what to do now.. Nov 14 '17 at 19:03
• Okay.. I think I did it.. $P(x)=4(x-3)-1 = 4x-13$ $P(5)=20-13 = 7$. This was a pretty confusing problem though.. Nov 14 '17 at 19:10
Hint 1: Show $\deg(P(P(x)))=\deg(P(x))^2$.
Hint 2: Solve $a(ax+b)+b=16x-5$