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$$P(x) = x^{2a+b-1} +x^{a-2b+5}-2x^{a+b-1}$$

This polynomial is divisible by $(x-2)$ so the remainder is $0$. How do we evaluate the product $ab$?

My attempt:

This polynomial is divisible by $(x-2)$ so the remainder is $0$

$$P(x) = (x-2)Q(x) + 0 \implies P(2) = 0$$

Plugging $x = 2$

$$P(2) = (2)^{2a+b-1}+(2)^{a-2b+5}-2(2)^{a+b-1}$$

Let $2^a = u$ and $2^b = m$

$$0 = \dfrac{1}{2}(u^2m)+32um^{-2}-2\dfrac{1}{2}(um) = \dfrac{1}{2}(u^2m)+32um^{-2}-um$$

Multiplying both sides by $2m^2$

$$0 = u^2m^3+64u-2um^3$$

Factoring $u$

$$0 = u\biggr(um^3+64-2m^3\biggr ) $$

Here we get one solution for $u = 0$. But I'm not sure If I went correctly.

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    $\begingroup$ $u$ can't possibly be $0$, so it must be that other term which is $0$. $\endgroup$ – Arthur Feb 27 '19 at 6:43
  • $\begingroup$ @Arthur Yes, right. We cannot get real solutions for $2^a = 0$. However, I do not also know what to do in this case. $\endgroup$ – Bobtrollsten Feb 27 '19 at 6:47
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We continue from the last line, since it has been pointed out in comments that $u\ne0$: $$0=um^3+64-2m^3=(u-2)m^3+64$$ $$64=(2-u)m^3$$ Given that $u,m$ are powers of 2, $2-u$ is positive, but $u>0$ so $u=1$ and $m=4$, i.e. $a=0,b=2$ and $ab=0$.

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  • $\begingroup$ Technically, we only know that $2a+b-1$, $a-2b+5$ and $a+b-1$ are natural numbers. Not $a$ and $b$. $\endgroup$ – Arthur Feb 27 '19 at 6:54
  • $\begingroup$ @Arthur As usual, the question has been set up so that the natural assumptions follow through nicely ;) $\endgroup$ – Parcly Taxel Feb 27 '19 at 6:55
  • $\begingroup$ Not exactly. When you say "Given that $u,m$ are powers of 2", you're implicitly assuming that $a$ and $b$ are non-negative integers, without having proven it. You then use this assumption when you conclude that $u=1$ from $0<u<2$. But you haven't proven that $a$ is a non-negative integer, so $u=1$ may not be true. $\endgroup$ – Arthur Feb 27 '19 at 8:37
  • $\begingroup$ @Arthur It all just works out in the end... $\endgroup$ – Parcly Taxel Feb 27 '19 at 10:58
  • $\begingroup$ It works out to a solution. You still don't know whether there are more. $\endgroup$ – Arthur Feb 27 '19 at 11:23

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