Originally Posted by

**Jonroberts74** let me reword,

the possible rational factors are $\displaystyle \pm 1, 2$ which means that I can try to divide $\displaystyle x^3-3x+2$ by the following $\displaystyle x-1, x+1, x+2,x-2$, but the the two that result in zero are $\displaystyle x-1, x+2$ which means $\displaystyle x=1$ or $\displaystyle x = -2$

That's not what I was trying to understand though,

I was reviewing an example and then asked someone else who confirmed it. In the polynomial the person divided by the factor 3, then they were able to divide the resulting polynomial by 3 again and then they ended up with a standard form polynomial which they further solved for two more solutions of x = -4 and x = 3. So they ended up with $\displaystyle (x-3)(x-3)(x-3)(+4) \Rightarrow (x-3)^3(x+4)$ as the factored version of the polynomial. So I was under the impression that I should have been able to divide $\displaystyle x^3-3x+2$ by (x-1) then the resulting polynomial by (x-1).

As I am replying I see that would be the case if the factor resulted in $\displaystyle (x-1)^3(x+2)$ but not for $\displaystyle (x-1)^2(x+2)$ because it leaves a polynomial in standard form--which already has two factors (when factored) and my answer results in three (x-1)(x-1)(x+2).