How do I find the partial fraction decomposition of (x^4+1)/(x^5+4x^3)x4+1x5+4x3 ?
1 Answer
First we will factor the denominator as much as possible:
(x^4 + 1)/(x^3(x^2 + 4))x4+1x3(x2+4)
And now, we will choose the factors to write:
(x^4 + 1)/(x^3(x^2 + 4)) = A/x + B/x^2 + C/x^3 + (Dx+E)/(x^2 + 4)x4+1x3(x2+4)=Ax+Bx2+Cx3+Dx+Ex2+4
Note that since there was a lone power of
The next step is to multiply both sides of the equation by
x^4 + 1 = Ax^2*(x^2 + 4) + Bx*(x^2+4) +x4+1=Ax2⋅(x2+4)+Bx⋅(x2+4)+
C*(x^2 + 4) + x^3(Dx+E)C⋅(x2+4)+x3(Dx+E)
And now, we will distribute and simplify everything:
x^4 + 1 = Ax^4 + 4Ax^2 + Bx^3 + 4Bx + Cx^2 +x4+1=Ax4+4Ax2+Bx3+4Bx+Cx2+
4C + Dx^4 + Ex^34C+Dx4+Ex3
We can solve for each constant now, by using the technique of grouping. The first step is to rearrange everything in successive powers of
x^4 + 1 = Ax^4 + Dx^4 + Bx^3 + Ex^3 + 4Ax^2 + Cx^2+x4+1=Ax4+Dx4+Bx3+Ex3+4Ax2+Cx2+
4Bx + 4C4Bx+4C
And now, we will factor out the constant terms:
x^4 + 1 =(A + D)x^4 + (B+E)x^3 + (4A+C)x^2 + 4Bx + 4Cx4+1=(A+D)x4+(B+E)x3+(4A+C)x2+4Bx+4C
The next step is to create a system of equations using the coefficients of
This implies that
Immediately from the last two equations, we can conclude that
From this it follows that since
Then, after plugging
Now all that's left is to plug these coefficient values into our expanded expression:
(x^4 + 1)/(x^3(x^2 + 4)) = A/x + B/x^2 + C/x^3 + (Dx+E)/(x^2 + 4)x4+1x3(x2+4)=Ax+Bx2+Cx3+Dx+Ex2+4
(x^4 + 1)/(x^3(x^2 + 4)) = 1/(4x^3) + (17x)/(16(x^2 + 4)) - 1/(16x)x4+1x3(x2+4)=14x3+17x16(x2+4)−116x
And there we have it. Remember, successfully expanding with partial fractions is all about choosing the correct factors, and from there it's just a lot of algebra. If you are familiar with the grouping technique, then you shouldn't have any trouble solving for the coefficients.
