- #1

BOAS

- 553

- 19

Hello,

i've come across a partial fractions problem that I don't know how to solve - Usually, the denominator of the fraction I need to split up into two separate fractions is a quadratic, but in this instance it's a cubic.

Specifically, the problem I'm having is that two of the factors to this cubic are the same, so when I set x equal to values in order to find the unknowns, everything disappears!

Write [itex]\frac{(x - 1)}{(x + 1)(x + 2)^{2}}[/itex] as the sum of 3 partial fractions.

[itex]\frac{(x - 1)}{(x + 1)(x + 2)^{2}} = \frac{A}{(x + 1)} + \frac{B}{(x + 2)} + \frac{C}{(x + 2)}[/itex]

[itex] = \frac{A(x + 2)(x + 2)}{(x + 1)(x + 2)(x + 2)} + \frac{B(x + 1)(x + 2)}{(x + 1)(x + 2)(x + 2)} + \frac{C(x + 1)(x + 2)}{(x + 1)(x + 2)(x + 2)}[/itex]

[itex](x - 1) \equiv A(x +2)(x + 2) + B(x+1)(x + 2) + C(x + 1)(x + 2)[/itex]

If I set x = -1, then A = -2

But now if I try to make the other brackets equal to zero, A, B and C disappear...

I'm probably missing something obvious (or shouldn't be approaching this in the exact same way I would a simpler one), but I can't see the solution.

Thanks for any help you can give

i've come across a partial fractions problem that I don't know how to solve - Usually, the denominator of the fraction I need to split up into two separate fractions is a quadratic, but in this instance it's a cubic.

Specifically, the problem I'm having is that two of the factors to this cubic are the same, so when I set x equal to values in order to find the unknowns, everything disappears!

## Homework Statement

Write [itex]\frac{(x - 1)}{(x + 1)(x + 2)^{2}}[/itex] as the sum of 3 partial fractions.

## Homework Equations

## The Attempt at a Solution

[itex]\frac{(x - 1)}{(x + 1)(x + 2)^{2}} = \frac{A}{(x + 1)} + \frac{B}{(x + 2)} + \frac{C}{(x + 2)}[/itex]

[itex] = \frac{A(x + 2)(x + 2)}{(x + 1)(x + 2)(x + 2)} + \frac{B(x + 1)(x + 2)}{(x + 1)(x + 2)(x + 2)} + \frac{C(x + 1)(x + 2)}{(x + 1)(x + 2)(x + 2)}[/itex]

[itex](x - 1) \equiv A(x +2)(x + 2) + B(x+1)(x + 2) + C(x + 1)(x + 2)[/itex]

If I set x = -1, then A = -2

But now if I try to make the other brackets equal to zero, A, B and C disappear...

I'm probably missing something obvious (or shouldn't be approaching this in the exact same way I would a simpler one), but I can't see the solution.

Thanks for any help you can give

Last edited: