Partial Fraction Help?

**Cbcollier91** · March 20th 2010, 01:15 PM

I need a little help with this problem:
8/(y^3-4y).
The book says the answer is -2/y+1/(y-2)+1/y+2 but I'm not following how they got that answer.
I have a feeling my main problem is that my algebra is rusty

.

**Archie Meade** · March 20th 2010, 01:22 PM

Originally Posted by Cbcollier91

I need a little help with this problem:
8/(y^3-4y).
The book says the answer is -2/y+1/(y-2)+1/y+2 but I'm not following how they got that answer.
I have a feeling my main problem is that my algebra is rusty

.

$\frac{8}{y^3-4y}=\frac{8}{y\left(y^2-4\right)}=\frac{8}{y(y+2)(y-2)}$

Then seperate this into

$\frac{8}{y(y+2)(y-2)}=\frac{A}{y}+\frac{B}{y+2}+\frac{C}{y-2}$

Lastly, find A, B and C

**skeeter** · March 20th 2010, 01:25 PM

Originally Posted by Cbcollier91

I need a little help with this problem:
8/(y^3-4y).
The book says the answer is -2/y+1/(y-2)+1/y+2 but I'm not following how they got that answer.
I have a feeling my main problem is that my algebra is rusty

.

$\frac{8}{y(y+2)(y-2)} = \frac{A}{y} + \frac{B}{y+2} + \frac{C}{y-2}$

$8 = A(y+2)(y-2) + By(y-2) + Cy(y+2)<br />$

let $y = 0$ ... $A = -2$

let $y = 2$ ... $C = 1$

let $y = -2$ ... $B = 1$

$\frac{8}{y(y+2)(y-2)} = -\frac{2}{y} + \frac{1}{y+2} + \frac{1}{y-2}$

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