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C4 Partial Fractions- help!

Write x3−3x2+5x−2(x+1)2\frac{x^{3} -3x^{2} +5x -2}{(x+1)^{2}} in the form x+A+Bx+1+C(x+1)2x +A +\frac{B}{x+1} +\frac{C}{(x+1)^{2}} where A, B, C and x are constants to be found.

So far, I've found C to be -11 and B to be 12, but B is apparently 14...
x3−3x2+5x−2=x(x+1)2+A(x+1)2+B(x+1)+Cx^{3} -3x^{2} +5x -2=x(x+1)^{2}+A(x+1)^{2}+B(x+1)+C
I subbed in C=-11 and then added that to both sides so I was left with only A and B terms on the RHS. I then subbed in x=0 and then x=1 to form two simultaneous equations from which I found that B was 12...Any ideas?
Reply 1
Avatar for Hasufel
Hasufel
16
the degree of the numerator is greater than the degree of the denominator, so first I`d divide out (you get an ax+b in doing this) this simplifies things with the remaining rational fraction, where you only have to use ascending powers of (x+1) (the line you have above is fine, and you`ll have what "A" is already...

equate x equal to -1 to get C, then choose an arbitrary x to get B

(EDIT) but, you can "guess" what B and C are by factorising the "remainder" i.e.

14x+3=14(x+1)−11=B(x+1)+C14x+3=14(x+1)-11=B(x+1)+C
(edited 10 years ago)
Original post by bobbricks
Write x3−3x2+5x−2(x+1)2\frac{x^{3} -3x^{2} +5x -2}{(x+1)^{2}} in the form x+A+Bx+1+C(x+1)2x +A +\frac{B}{x+1} +\frac{C}{(x+1)^{2}} where A, B, C and x are constants to be found.

So far, I've found C to be -11 and B to be 12, but B is apparently 14...
x3−3x2+5x−2=x(x+1)2+A(x+1)2+B(x+1)+Cx^{3} -3x^{2} +5x -2=x(x+1)^{2}+A(x+1)^{2}+B(x+1)+C
I subbed in C=-11 and then added that to both sides so I was left with only A and B terms on the RHS. I then subbed in x=0 and then x=1 to form two simultaneous equations from which I found that B was 12...Any ideas?


Comparing the coefficients of x^2 will immediately give you the value of A. Then coefficients of x will give you B
Reply 3
Avatar for bobbricks
bobbricks
OP
19
Original post by Hasufel
the degree of the numerator is greater than the degree of the denominator, so first I`d divide out (you get an ax+b in doing this) this simplifies things with the remaining rational fraction, where you only have to use ascending powers of (x+1) (the line you have above is fine, and you`ll have what "A" is already...

equate x equal to -1 to get C, then choose an arbitrary x to get B

(EDIT) but, you can "guess" what B and C are by factorising the "remainder" i.e.

14x+3=14(x+1)−11=B(x+1)+C14x+3=14(x+1)-11=B(x+1)+C



Original post by brianeverit
Comparing the coefficients of x^2 will immediately give you the value of A. Then coefficients of x will give you B

Thanks you two, I've finally managed to do it! :biggrin: :biggrin:

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