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C4 binomial help

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Hey just doing some Easter revision. I have no idea why my answer to the second part of this question is horrifically wrong

Any help would be greatly appreciated, thanks
Reply 1
Original post by hi-zen-berg


Hey just doing some Easter revision. I have no idea why my answer to the second part of this question is horrifically wrong

Any help would be greatly appreciated, thanks


4(25x)1=4(2(15/2x))1=4(21)(15/2x)=2(15/2x)4(2 - 5x)^{-1} = 4(2(1-5/2x))^{-1} = 4(2^{-1})(1-5/2x) = 2(1 - 5/2x)

Then you cube that. You've multiplied by 2 instead.
(edited 8 years ago)
Original post by Zacken
4(25x)1=4(2(15/2x))1=4(21)(15/2x)=2(15/2x)4(2 - 5x)^{-1} = 4(2(1-5/2x))^{-1} = 4(2^{-1})(1-5/2x) = 2(1 - 5/2x)

Then you cube that. You've multiplied by 2 instead.


Thanks a lot man really helpful
Reply 3
Original post by hi-zen-berg
Thanks a lot man really helpful


Pleasure. :-)

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