Question 76000
You can have an infinite number of possible factors for the missing factor. For instance, we can have 

{{{(x+5)(2x-1)(x)}}}
or 
{{{(x+5)(2x-1)(x+2)}}} or
{{{(x+5)(2x-1)(x+22)}}} etc
So we can continue this forever. 

If we have the condition of one zero being x=1 then our factor must be 
{{{x-1=0}}} since if we have a zero of x=a we have a factor: {{{x-a=0}}}
So we would have only one possible polynomial
{{{(x+5)(2x-1)(x-1)}}}