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

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

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.

• $u$ can't possibly be $0$, so it must be that other term which is $0$. – Arthur Feb 27 '19 at 6:43
• @Arthur Yes, right. We cannot get real solutions for $2^a = 0$. However, I do not also know what to do in this case. – Bobtrollsten Feb 27 '19 at 6:47

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 $$ab=0$$.
• Technically, we only know that $2a+b-1$, $a-2b+5$ and $a+b-1$ are natural numbers. Not $a$ and $b$. – Arthur Feb 27 '19 at 6:54
• 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. – Arthur Feb 27 '19 at 8:37