How do you express as a partial fraction # (6x^2 - 3x +1) / ( (4x+1)(x^2 +1) )#?

Question

Paisley Johnson · Accepted Answer

# (6x^2 - 3x +1) / ( (4x+1)(x^2 +1) ) = 2/(4x+1)+(x-1)/(x^2+1)#

Start with: # (6x^2 - 3x +1) / ( (4x+1)(x^2 +1) )#

The denominator is already factored into irreducible polynomials (over the Reals).

Because #x^2+1# cannnot be factored using Real number coefficients, we need a linear numerator for one of the fractions

We need:

#A/(4x+1)+(Bx+C)/(x^2+1) = (6x^2 - 3x +1) / ( (4x+1)(x^2 +1) )#

This lead to: #(Ax^2+A+4Bx^2+4Cx+Bx+C) / ( (4x+1)(x^2 +1) )=(6x^2 - 3x +1) / ( (4x+1)(x^2 +1) )#

And so:

#((A+4B)x^2+(B+4C)x+(A+C)) / ( (4x+1)(x^2 +1) )=(6x^2 - 3x +1) / ( (4x+1)(x^2 +1) )#

So we need to solve the system:

#A# #+4B# #" " " " # # =# #6#

#" " " " # #B# #+4C# # =# #-3#

#A# #" " " " " # # +C# #=# #1#

Eq3 implies #A = 1-C# and substituting in Eq1 and simplifying gets us:

#" " " " # #4B# #-C# # =# #5#

Eq2 is #" " " " # #B# #+4C# # =# #-3#

So #" " " "# #-4B# #-16C# # =# #12#

Thus #-17C = 17# and #C= -1#

Knowing #C#, we can find #A = 1-(-1) =2#

And #B+4(-1) = -3# gets us #B = 1#

#A/(4x+1)+(Bx+C)/(x^2+1) = 2/(4x+1)+(x-1)/(x^2+1)#