If it has any rational solutions the numerator must be
a factor of 4 and the denominator must be a factor of 2,

The only possible rational solutions are 

, , , or 

Trying the easiest one x = 1

So we divide by x - 1, either synthetically:

1|2  11   1  -10  -4
 |    2  13   14   4 
  2  13  14    4   0

or by long division   

             2x³ + 13x² + 14x + 4
x - 1)2x4 + 11x³ +   x² - 10x - 4 
      2x4 -  2x³
            13x³ +   x²
            13x³ - 13x²
                   14x² - 10x
                   14x² - 14x
                           4x - 4
                           4x - 4
                                0     

You have now factored the original polynomial as

(x - 1)(2x³ + 13x² + 14x + 4)

Next let's do the same with the cubic polynomial in
the second parenthesis:

If it has any rational solutions the numerator must be
a factor of 4 and the denominator must be a factor of 2,

We've already done that. The only possible rational solutions are 

, , , or 

You could go to the trouble of trying all those.  But sooner of
later you'd get around to trying to divide by  or 

Divide by x + .5, either synthetically:

-.5|2  13  14   4
   |   -1  -6  -4    
    2  12   8   0

or by long division   

               2x² + 12x + 8 
x + .5)2x³ + 13x² + 14x + 4 
       2x³ +   x²
             12x² + 14x
             12x? +  6x
                     8x + 4
                     8x + 4
                          0         

And you have now factored further as

(x - 1)(x + .5)(2x² + 12x + 8)

Now we can factor 2 out of the third parentheses:

(x - 1)(x + .5)2(x² + 6x + 4)

2(x - 1)(x + .5)(x² + 6x + 4)

The quadratic will not factor but we can find its roots 
by using the quadratic formula:



















So the 4 roots are

1, , , and 

Edwin