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

Question

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

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.

Answer 1

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 $ab=0$.