# Partial Fraction Decomposition

• Feb 2nd 2011, 11:22 PM
dagbayani481
Partial Fraction Decomposition
Partial Fraction Decomposition of the Following:

$\displaystyle (4x^2-8x+48)/(x^4-8x^3+16x^2)$
I factored out and simplified the denominator which gives me: $\displaystyle x^2(x-4)(x-4)$.

From that point on I write out the decomposition relative to A,B,C,D and the factored out denominator which is: $\displaystyle (A/x^2)+(B/x)+(C/x-4)+(D/x-4)$.

I need to find the constants A B C & D that equal the original numerator.
In other words:
$\displaystyle (A/x^2)+(B/x)+(C/x-4)+(D/x-4)= (4x^2-8x+48)/(x^4-8x^3+16x^2)$

from the decomposition relative to A B C and D, I combine all terms under one single denominator and get:

$\displaystyle (Ax^2-8Ax+16A+Bx^3-8Bx^2+16Bx+Cx^3-4Cx^2+Dx^3-4Dx^2)/x^2(x^2-8x+16)$

I create systems of equations according to the degree of each variable and get:

$\displaystyle B+D+C=0$

$\displaystyle A-8B-4C-4D=4$

$\displaystyle -8A+16B=-8$

$\displaystyle 16A=48$

When I solve the system I get no solutions?
Can anyone tell me where I am going wrong?
• Feb 2nd 2011, 11:45 PM
From that point on I write out the decomposition relative to A,B,C,D and the factored out denominator which is: $\displaystyle (A/x^2)+(B/x)+(C/x-4)+(D/x-4)$.
the decomposition should be $\displaystyle \frac{4x^2-8x+48}{x^4-8x+16x^2}=\frac{A}{x^2}+\frac{B}{x}+\frac{C}{(x-4)}+\frac{D}{(x-4)^2}$