Find all the polynomials $p(x)$ with real coefficients such that :- $(x-8)P(2x)=8(x-1)P(x)$
I tried putting different values to P(x) to find any roots, here's what I found out.
$P(2)=P(4)=P(8)=0$
So,
$P(x)=(x-2)(x-4)(x-8)Q(x)$, Where $Q(x)$ is some polynomial.
Now I don't know what to do next.
Any help would be appreciated.
Certainly the zero polynomial is a solution. Let $P(x)=\sum_{k=0}^na_kx^k$ be a polynomial of degree $n$ (thus $a_n \neq 0$). Then the given relation suggests $$(x-8)\left(\sum_{k=0}^na_k(2x)^k\right)=8(x-1)\left(\sum_{k=0}^na_kx^k\right).$$ Compare the coefficient of $x^{n+1}$ on both sides to get $$a_n2^n=8a_n.$$ This implies $n=3$. Thus the polynomial $P$ (if not the zero polynomial) should have degree $3$. I think you have made an error in claiming that $P(1)=0$. Thus the only roots you have are $x=2,4,8$. This means $$P(x)=C(x-2)(x-4)(x-8).$$
