sum of partial fractions

[lotmasta]

Hi
I'm having some trouble figuring out this problem:
x^3-x^2
----------
(x+3)^4

I can write it like
...A..........B..............C...........D...............................
----...+.---------..+ ---------...+.---------.... =...x^3-x^2
x+3......(x+3)^2....(x+3)^3..(x+3)^4........... ----------
....................................................................(x+3)^4

And then multiplying everything by (x+3)^4

x^3-x^2 = A (x+3)^3 + B (x+3)^2 + C (x+3) + D

I can find D = -36.. but I'm a little stuck on what to do next since all the terms can be set to 0 if x = -3...

If someone could tell me how to get A B or C I think I could figure out the rest

Thanks

 

[Gene]

If you expand A*(x+3) you get the only x^3 term on the right, A*x^3. That has to equal the x^3 term on the other side so A = 1.

 

[soroban]

Hello, lotmasta!

I'm having some trouble figuring out this problem:

. . . .x³ - x²
. . . ----------
. . . (x + 3)⁴

I can write it like:

. . . . .A . . . . .B . . . . . . C . . . . . .D . . . . . . .x³ - x²
. . . ------ + -------- + -------- + --------- . = . ---------
. . . x + 3 . .(x+3)² . .(x+3)³ . . (x+3)⁴ . . . . (x+3)⁴


And then multiplying everything by (x+3)⁴:

. . . x³ - x² . = . A(x+3)³ + B(x+3)² + C(x+3) + D

I can find D = -36 . . . but I'm a little stuck on what to do next
since all the terms are set to 0 if x = -3.

If someone could tell me how to get A B or C I think I could figure out the rest.

This difficulty arises when we have <u>repeated</u> factors.

To solve for four variables, we need four equations.
. . . You already found one and got D = -36.
We need three more, so we plug in <u>any</u> three values for x.

Plug in x = 1: . 1³ - 1² .= .A(1+3)³ + B(1+3)² + C(1+3) + D
. . . and we get: . 64A + 16B + 4C + D .= .0 . [1]

Plug in x = -1, and we get: . 8A + 4B + 2C + D .= .-2 . [2]

Plug in x = 0, and we get: . 27A + 9B + 3C + D .= .0 . [3]


We already know that D = -36.
Substitute that into [1], [2], [3] and we get three new equations.

. . . [1'] . 64A + 16B + 4C .= .36 . ---> . 16A + 4B + C .= . 9
. . . [2'] . . 8A + . 4B + 2C .= .34 . ---> . . 4A + 2B + C .= .17
. . . [3'] . 27A + . 9B + 3C .= .36 . ---> . . 9A + 3B + C .= .12

Then solve this system of equations.
. . . A = 1 . . . B = -10 . . . C = 33