I'm not sure I understand your question. Can you tell us in what context you're doing this problem? I remember in calculus/diff. eq. we had partial fractions.
If you are doing this in order to integrate parts with partial fractions, then try this:
(16x^4) - 1 = (2x+1) (2x-1) ((4x^2) + 1)
x/((16x^4)- 1) = [A/(2x+1)] + [B/(2x-1)] + [(Cx+D)/((4x^2)+ 1)]
=> x = A(2x-1)[(4x^2)+ 1] + B(2x+1)[(4x^2)+ 1] + (Cx+D)(2x+1)(2x-1)
You can choose to use the linear system to find A, B, C and D; or you can choose any x that cancels out the other equations.
I will use the faster method:
If we let x = 1/2, then 1/2 = 0 + B(2*2) + 0 => 1/2 = 4B => B = 1/8
If we let x = -1/2, then -1/2 = A(-2*2) + 0 + 0 => -1/2 = -4A => A = 1/8
Let x = 0, then 0 = A(-1) + B(1) + (0+D)(-1) = -A + B -D = (-1/8) + (1/8) - D
=> -D = 0 => D = 0
We know the values of A, B and D. Take each value and replace, and choose any x that doesn't cancel out C to find C.
If you take x = 1, then 1 = 5A + B(3*5) + (C+D)*3 = (5/8) + (15/8) + 3C
=> 3C = 1 - (20/8) = -12/8 => C = -1/2
As you can see, you can quickly know what A, B and D are, but C requires more work. It could look messy.
The second method is much neater. You expand the equation, then use the a matrix to solve it (The second method):
x = (8Ax^3 + 2Ax - 4Ax^2 - A) + (8Bx^3 + 2Bx + 4Bx^2 + B) + (4Cx^3 - Cx + 4Dx^2 - D)
= (8A + 8B + 4C) x^3 + (-4A + 4B + 4D) x^2 + (2A + 2B - C)x + (-A + B -D)
So,
8A + 8B + 4C = 0
-4 A + 4B + 4D = 0
2A + 2B - C = 1
-A + B - D = 0
You can make a matrix and find the solution. But since I have already gotten the values of A and B, I can just plug them in and find C and D (supposing I had difficulty with the two).
4C = -8A - 8B = -1 -1 = -2 => C = -1/2
D = -A + B = -1 + 1 => D = 0
Hence, A = 1/8; B= 1/8; C= -1/2; D = 0
Hope that helps! I may have made an error somewhere, so check!
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