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partial fractions

(x+4)/(x-1)^3

x+4= A(x-1)^2+B(x-1)+C
I sub in x=1 and i get C=5 but how do i work out A and B. When i sub in x=0 , i get 4= A - B
Reply 1
Original post by man111111
(x+4)/(x-1)^3

x+4= A(x-1)^2+B(x-1)+C
I sub in x=1 and i get C=5 but how do i work out A and B. When i sub in x=0 , i get 4= A - B

You could sub in another value of x to get a second equation and then solve simultaneously.

A better way is to expand the right-hand-side and compare coefficients. Although without any expanding you should be able to see what A must be.
(edited 6 years ago)
Reply 2
thanks
Original post by man111111
(x+4)/(x-1)^3

x+4= A(x-1)^2+B(x-1)+C
I sub in x=1 and i get C=5 but how do i work out A and B. When i sub in x=0 , i get 4= A - B


To add to what notnek said, alternatively you can do this question without any work whatsoever by simply adding 0 creatively. More specifically,

x+4(x1)3=(x1)+5(x1)3=1(x1)2+5(x1)3\dfrac{x+4}{(x-1)^3}=\dfrac{(x-1)+5}{(x-1)^3}=\dfrac{1}{(x-1)^2}+\dfrac{5}{(x-1)^3}
(edited 6 years ago)
Reply 4
Original post by RDKGames
To add to what notnek said, alternatively you can do this question without any work whatsoever by simply adding 0 creatively. More specifically,

x+4(x1)3=(x1)+5(x1)3=1(x1)2+5(x1)3\dfrac{x+4}{(x-1)^3}=\dfrac{(x-1)+5}{(x-1)^3}=\dfrac{1}{(x-1)^2}+\dfrac{5}{(x-1)^3}

I wish this technique was taught in schools more and part of A Level textbooks. It probably requires a textbook exercise for a student to become confident with it.
Original post by Notnek
I wish this technique was taught in schools more and part of A Level textbooks. It probably requires a textbook exercise for a student to become confident with it.


Yes it definitely cuts workings short in most PF questions, and isn’t that difficult to get used to with little practice.
Do you teach it to your students regardless of not being part of spec/textbooks?
Reply 6
Original post by RDKGames
To add to what notnek said, alternatively you can do this question without any work whatsoever by simply adding 0 creatively. More specifically,

x+4(x1)3=(x1)+5(x1)3=1(x1)2+5(x1)3\dfrac{x+4}{(x-1)^3}=\dfrac{(x-1)+5}{(x-1)^3}=\dfrac{1}{(x-1)^2}+\dfrac{5}{(x-1)^3}


thanks by the way i was wondering if you were given 2x +2 dx with limits, can you write it as 2x + 2 dx with limits on each integral sign? or are you only able to write it like this if it didnt have any limits
Original post by man111111
thanks by the way i was wondering if you were given 2x +2 dx with limits, can you write it as 2x + 2 dx with limits on each integral sign? or are you only able to write it like this if it didnt have any limits


The splitting is a propety of integrals, so yes you can as long as the limits are the same. You can also do it on indefinite integrals too.
Reply 8
Original post by RDKGames
The splitting is a propety of integrals, so yes you can as long as the limits are the same. You can also do it on indefinite integrals too.


thanks
Reply 9
Original post by RDKGames
Yes it definitely cuts workings short in most PF questions, and isn’t that difficult to get used to with little practice.
Do you teach it to your students regardless of not being part of spec/textbooks?

That depends on time :smile: For further maths students I would always teach it.
Reply 10
Original post by RDKGames
The splitting is a propety of integrals, so yes you can as long as the limits are the same. You can also do it on indefinite integrals too.


how do i work out x(2x+1)/ (x+1). i cant seem to make it into a proper fraction
Reply 11
never mind, i was dividing it incorrectly
Original post by man111111
how do i work out x(2x+1)/ (x+1). i cant seem to make it into a proper fraction


You can use long division to determine the partial fractions. Otherwise, I can tell you it will be in the form Ax+B+Cx+1Ax+B+\dfrac{C}{x+1}

Alternatively, use the same trick as I've shown above, x(2x+1)x+1=2x(x+1)xx+1=2x(x+1)1x+1\dfrac{x(2x+1)}{x+1} = \dfrac{2x(x+1)-x}{x+1} = 2x-\dfrac{(x+1)-1}{x+1} .....
Original post by RDKGames
You can use long division to determine the partial fractions. Otherwise, I can tell you it will be in the form Ax+B+Cx+1Ax+B+\dfrac{C}{x+1}

Alternatively, use the same trick as I've shown above, x(2x+1)x+1=2x(x+1)xx+1=2x(x+1)1x+1\dfrac{x(2x+1)}{x+1} = \dfrac{2x(x+1)-x}{x+1} = 2x-\dfrac{(x+1)-1}{x+1} .....


What....
Original post by Bulletzone
What....


What? It's basic bracket manipulation
Reply 15
Original post by RDKGames
You can use long division to determine the partial fractions. Otherwise, I can tell you it will be in the form Ax+B+Cx+1Ax+B+\dfrac{C}{x+1}

Alternatively, use the same trick as I've shown above, x(2x+1)x+1=2x(x+1)xx+1=2x(x+1)1x+1\dfrac{x(2x+1)}{x+1} = \dfrac{2x(x+1)-x}{x+1} = 2x-\dfrac{(x+1)-1}{x+1} .....


thanks do you mind telling me your thought process of getting x(2x+1) = 2x(x+1) - x
Original post by man111111
thanks do you mind telling me your thought process of getting x(2x+1) = 2x(x+1) - x


Well we have a x+1 in the denominator so we would like to get this somehow on the numerator in order for them to cancel. So, you can notice that 2x+1 is almost 2(x+1). We can get it by saying that 2x+1 = 2x+2-1 hence we get x(2x+1) = x(2x+2-1) = x(2x+2)-x = 2x(x+1) - x
Reply 17
Original post by RDKGames
Well we have a x+1 in the denominator so we would like to get this somehow on the numerator in order for them to cancel. So, you can notice that 2x+1 is almost 2(x+1). We can get it by saying that 2x+1 = 2x+2-1 hence we get x(2x+1) = x(2x+2-1) = x(2x+2)-x = 2x(x+1) - x


thanks

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