Find a polynomial from an equality

Question

Find all polynomials for which

What I have done so far: for $x=8$ we get $p(8)=0$ for $x=1$ we get $p(2)=0$

So there exists a polynomial $p(x) = (x-2)(x-8)q(x)$

This is where I get stuck. How do I continue?

UPDATE

After substituting and simplifying I get $(x-4)(2ax+b)=4(x-2)(ax+b)$

For $x = 2,8$ I get

$x= 2 \to -8a+b=0$

$x= 8 \to 32a+5b=0$

which gives $a$ and $b$ equal to zero.

Answer 1

The route you take is fruitful.

$p\left(x\right)=\left(x-2\right)\left(x-8\right)q\left(x\right)$ leads to:

$$\left(x-4\right)q\left(2x\right)=2\left(x-2\right)q\left(x\right)$$

Then $4$ must be a root of $q$, so $q\left(x\right)=\left(x-4\right)r\left(x\right)$ leading to:

$$r\left(2x\right)=r\left(x\right)$$

Then $r\left(x\right)$ must be a constant polynomial and we end up with: $$p\left(x\right)=c\left(x-2\right)\left(x-4\right)\left(x-8\right)$$

Answer 2

HINT:

Let the highest of power of $x$ be $n$

So, $(x-8)[a(2x)^n+\cdots]=8(x-1)[ax^n+\cdots]$

Comparing the coefficients of $x^{n+1},$ $$a2^n=8a\implies n=3$$

Let $p(x)=(x-2)(x-8)(ax+b)$ where $a,b$ are arbitrary constants to be determined

Hope you take it from here?

Answer 3

The following is essentially @drhab's solution, but uses only one idea repeatedly.

From $$ (x-8)p(2x) = 8(x-1)p(x) $$ we see $x-8$ divides $p(x)$. Let $p(x) = (x-8)p_1(x)$ and substitute, yielding $$ 2(x-8)(x-4)p_1(2x) = 8(x-1)(x-8)p_1(x) $$ From this we see $x-4$ divides $p_1(x)$. Let $p_1(x) = (x-4)p_2(x)$ and substitute, yielding $$ 4(x-8)(x-4)(x-2)p_2(2x) = 8(x-1)(x-4)(x-8)p_2(x) $$ From this we see $x-2$ divides $p_2(x)$. Let $p_2(x) = (x-2)p_3(x)$ and substitute, yielding $$ 8(x-8)(x-4)(x-2)(x-1)p_3(2x) = 8(x-1)(x-2)(x-4)(x-8)p_3(x) $$ ( ... and our recursive process stops because the new $x-1$ factor divides the $x-1$ that's been lingering on the right all along.) But now we simplify to $p_3(2x) = p_3(x)$ and the rest of @drhab's argument finishes the argument.

Answer 4

Following the method outlined in this answer we can write the original equation in the form $$\frac{\sigma p}{p} = \frac{\sigma^3 r}{r}$$ where $\sigma p(x) = p(2x)$ and $r(x)=8-x$. Using the "additive notation" (see the referenced post) we obtain $$p=\frac{\sigma^3-1}{\sigma-1}r=(\sigma^2+\sigma+1)r=(4x-8)(2x-8)(x-8)$$ unique up to a constant factor.