Since the numerator has a higher degree than the denominator, before we can do
partial fractions, we must first do the long division to get a rational
expression which has a numerator of lower degree than the denominator:

                                        x + 0                                
2x³ + 5x² + 8x + 3)2x4 + 5x³ + 11x² - 15x - 3
                   2x4 + 5x³ +  8x² +  3x
                         0x³ +  3x² - 18x - 3
                         0x³ +  0x² +  0x + 0
                                3x² - 18x - 3

 = x + 

We'll break  into partial fractions.  Then
we'll come back and add the x quotient to it.

Now we factor the denominator.  We are told that 2x+1 is a factor, so we
divide that into the denominator

              x² + 2x + 3      
2x + 1)2x³ + 5x² + 8x + 3
       2x³ +  x²
             4x² + 8x
             4x² + 2x
                   6x + 3
                   6x + 3 

x²+2x+3 does not factor, so we have 

 = 

to break into partial fractions:


 =  + 

Multiply through by the LCD

3x² - 18x - 3 = A(x² + 2x + 3) + (Bx + C)(2x + 1)

This must be true for all values of x, so we substitute various values for x:

Substituting x=0

3(0)² - 18(0) - 3 = A((0)² + 2(0) + 3) + (B(0) + C)(2(0) + 1)

               -3 = 3A + C 

Substituting x=1

3(1)² - 18(1) - 3 = A((1)² + 2(1) + 3) + (B(1) + C)(2(1) + 1)
       3 - 18 - 3 = A(1 + 2 + 3) + (B + C)(2 + 1)
              -18 = A(6) + (B + C)(3)
              -18 = 6A + 3B + 3C

Substituting x=-1

3(-1)² - 18(-1) - 3 = A((-1)² + 2(-1) + 3) + (B(-1) + C)(2(-1) + 1)
       3 + 18 - 3 = A(1 - 2 + 3) + (-B + C)(-2 + 1)
               18 = A(2) + (-B + C)(-1)
               18 = 2A + B - C

So we solve the system of three equations:

-3 = 3A + C
-18 = 6A + 3B + 3C
18 = 2A + B - C

and get A=3, B=0, C=-12

So

 =  + 
 
  =  - 

Therefore going back to the original problem:

 = x +  - 

Edwin