The Student Room Group

Partial fractions in binomial expansion

By using partial fractions, or otherwise, find the first three terms in the expansion of
1 / (x^2-3x-4)

Ended up with 1/5(x-4) - 1/5(x+1)
But unsure if i did the partial fractions bit right..
(Missed half the lessons on this topic and now the teacher's away..)

I assume you take the 1/5 out but how would you go about expanding that? (or more likely i've gone wrong from the start ngl)

Many thanks:biggrin:
Original post by okcherubble88
By using partial fractions, or otherwise, find the first three terms in the expansion of
1 / (x^2-3x-4)

Ended up with 1/5(x-4) - 1/5(x+1)
But unsure if i did the partial fractions bit right..
(Missed half the lessons on this topic and now the teacher's away..)

I assume you take the 1/5 out but how would you go about expanding that? (or more likely i've gone wrong from the start ngl)

Many thanks:biggrin:


The partials are fine

Binomial (x+1)^-1 and divide by 5

Do the same for the other bracket

Do the subtraction
Original post by TenOfThem
The partials are fine

Binomial (x+1)^-1 and divide by 5

Do the same for the other bracket

Do the subtraction


Thanks!
So for the (x-4) bracket does it have to be divided by 4 to get x/4 -1?
Original post by okcherubble88
Thanks!
So for the (x-4) bracket does it have to be divided by 4 to get x/4 -1?


(x4)1=14(1x4)1(x-4)^{-1} = -\frac{1}{4}(1-\frac{x}{4})^{-1}
Original post by TenOfThem
(x4)1=14(1x4)1(x-4)^{-1} = -\frac{1}{4}(1-\frac{x}{4})^{-1}


Cheers :smile: should be fine now!

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