Partial fractions - alternative result I have the following fraction: z (

Kendrick Hampton

Kendrick Hampton

Answered question

2022-06-08

Partial fractions - alternative result
I have the following fraction:
z ( z 1 ) ( z 2 )
When I try to decompose it to partial fractions, I get:
1 z 1 + 2 z 2
But the result in my book is:
z z 1 + z z 2
Both results are correct, but, how am I supposed to get the second one?

Answer & Explanation

hildiadau0o

hildiadau0o

Beginner2022-06-09Added 21 answers

I agree with the comments that the result stated in the back of the book is somewhat non-standard. The way that you can see that the two results are the same is by cleverly adding and subtracting 1.
1 z 1 + 2 z 2 = 1 z 1 1 + 2 z 2 + 1 = 1 z 1 ( z 1 ) z 1 + 2 z 2 + z 2 z 2 = z z 1 + z z 2
arridsd9

arridsd9

Beginner2022-06-10Added 12 answers

z ( z 1 ) ( z 2 ) = z ( z 1 ) z ( z 2 ) ( z 1 ) ( z 2 ) = . . .

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