Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

c4: quick question re: partial fractions

Hi,

If I have a fraction, say

\frac{1}{u(u-1)}

because I have a squared term in the denominator, would the fraction be equivalent to

\frac{A}{u} + \frac{B}{u^2} - \frac{C}{u}

?

Is the above the best way (only way?) to do it?

Thanks

Original post by marcsaccount

Hi,

If I have a fraction, say

\frac{1}{u(u-1)}

because I have a squared term in the denominator, would the fraction be equivalent to

\frac{A}{u} + \frac{B}{u^2} - \frac{C}{u}

?

Is the above the best way (only way?) to do it?

Thanks

That doesn't give you the right denominator. You only have linear terms so the partial fractions are

\frac{A}{u} + \frac{B}{u-1}

If we combine your terms we get

\frac{A}{u} + \frac{B}{u^2} - \frac{C}{u} = \frac{D}{u} + \frac{B}{u^2} = \frac{Du+B}{u^2}

where

D=A-B

OP

Original post by atsruser

That doesn't give you the right denominator. You only have linear terms so the partial fractions are

\frac{A}{u} + \frac{B}{u-1}

If we combine your terms we get

\frac{A}{u} + \frac{B}{u^2} - \frac{C}{u} = \frac{D}{u} + \frac{B}{u^2} = \frac{Du+B}{u^2}

where

D=A-B

Hi - thanks for your help. I thought it was a repeated term because if the bracket is expanded the denominator would be

u^2 - u

, but this isn't a repeated term?

It would be repeated if it was

u(u-1)^2

as you see the

(u-1)

is repeated twice due to the second order.

Original post by marcsaccount

Hi - thanks for your help. I thought it was a repeated term because if the bracket is expanded the denominator would be

u^2 - u

, but this isn't a repeated term?

You are looking for terms that are multiplied in the denominator, not added or subtracted, due to the way fractions with different denominators are added - think about finding a common denominator when adding fractions:

\frac{a}{x} + \frac{b}{y} = \frac{ay}{xy} + \frac{xb}{xy} = \frac{ay+bx}{xy}

Note that we multiplied, not added, the denominators.

OP

OK, it's clear now, thanks for your help :smile:

:smile:

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