Writing $\frac{x^4(1-x)^4}{1+x^2}$ in terms of partial fractions

Question

How does one write

$$\frac{x^4(1-x)^4}{1+x^2}$$

in terms of partial fractions?

My Attempt

$$\frac{x^4(1-x)^4}{1+x^2}=\frac{x^4-4x^5+6x^6-4x^7+x^8}{1+x^2}$$

$$=Ax^6+Bx^5+Cx^4+Dx^3+Ex^2+Fx+\frac{G}{1+x^2}$$

Multiplying out and comparing coefficient gives:

$$=x^6-4x^5+5x^4-4x^2+0+\frac{0}{1+x^2}$$

This is obviously not correct, how to do this properly?

Answer 1

You should have tried $$\frac{x^4(1-x)^4}{1+x^2}=Ax^6+Bx^5+Cx^4+Dx^3+Ex^2+Fx +\mathbf{J}+\frac{\mathbf{G + Hx}}{1+x^2}$$

By the way, multiplying out and comparing coefficients gives: $$\frac{x^4(1-x)^4}{1+x^2}=x^6-4x^5+5x^4-4x^2+\mathbf{4}+\frac{\mathbf{-4}}{1+x^2}$$

Answer 2

To answer your comment about long division: you want to divide $x^8-4x^7+6x^6-4x^5+x^4$ by $x^2+1$. Then,

$$x^8-4x^7+6x^6=x^6(x^2+1)-4x^7+5x^6$$ $$-4x^7+5x^6-4x^5=-4x^5(x^2+1)+5x^6$$ $$5x^6+x^4=5x^4(x^2+1)-4x^4$$ $$-4x^4=-4x^2(x^2+1)+4x^2$$ $$4x^2=4(x^2+1)-4$$

All in all,

$$x^8-4x^7+6x^6-4x^5+x^4=(x^6-4x^5+5x^4-4x^2+4)(x^2+1)-4$$

Or

$$\frac{x^8-4x^7+6x^6-4x^5+x^4}{x^2+1}=x^6-4x^5+5x^4-4x^2+4-\frac4{x^2+1}$$

Usually you write it like this

First step:

x^8 -4x^7 +6x^6 -4x^5 +x^4    | x^2 + 0x + 1
    -4x^7 +5x^6 -4x^5 +x^4    +-----------------------
                              | x^6

Second step:

x^8 -4x^7 +6x^6 -4x^5 +x^4    | x^2 + 0x + 1
    -4x^7 +5x^6 -4x^5 +x^4    +-----------------------
          +5x^6 +0x^5 +x^4    | x^6 -4x^5

Then, the complete division:

x^8 -4x^7 +6x^6 -4x^5 +x^4            | x^2 + 0x + 1
    -4x^7 +5x^6 -4x^5 +x^4            +-----------------------
          +5x^6 +0x^5 +x^4            | x^6 -4x^5 +5x^4 -4x^2 +4
                     -4x^4            |
                            4x^2      |
                                 -4   |

It's not very easy to write or to read on computer, so feel free to ask if something is not clear enough.