All candidates for rational solutions must have a numerator which
divides evenly into the constant term in absolute value |-21|=21;
and whose denominator divides evenly into the absolute value of 
the leading coefficient |2|=2.

candidates for numerators of rational solutions: 1,3,7,21
 
candidates for denominators of rational solutions: 1,2

candidates for rational solutions: 


1/1, 1/2, 3/1, 3/2, 7/1, 7/2, 21/1, 21/2

or

1, 1/2, 3, 3/2, 7, 7/2, 21, 21/2



Try 1, using synthetic division:

1 | 2  -5 -17  41 -21
  |     2  -3 -20  21 
    2  -3 -20  21   0

That factors the left side as



The 0 on the bottom right tells us that 1 is a rational solution


Try 1 again in the quotient because it might have multiplicity 
more than 1.

1 | 2  -3 -20  21 
  |     2  -1 -21  
    2  -1 -21   0

That factors the left side again as



Again the 0 on the bottom right tells us that 1 is a second solution
of at least multiplicity 2.
or



We already know how to finish factoring, for it is a quadratic:



x-1=0;  2x-7=0;  x+3=0
  x=1;    2x=7;    x=-3
           x=7/2

So: 
1 is a rational solution with multiplicity 2.
7/2 is a rational solution with multiplicity of multiplicity 1.
-3 is a rational solution with multiplicity of multiplicity 1.

Edwin