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need help on a c4 partial fractions please!

(-10x^2+17x-15)/2(2x-1)^2(x+4)

how could I separate the terms? i.e. A/? + B/? etc

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Reply 1
You need A2x1+B(2x1)2+Cx+4\frac{A}{2x-1}+\frac{B}{(2x-1)^2}+\frac{C}{x+4}
Reply 2
but it's also divided by 2
Reply 3
Original post by ahyak
but it's also divided by 2


The A,B and C values will account for that.
Reply 4
oh right, okay cheers :smile:
Just because I have this typed up from yesterday...

Here is an example:

2x2+4(x2)2(x+4)A(x2)+B(x2)2+Cx+4\frac{2x^2+4}{(x-2)^2(x+4)} \equiv \frac{A}{(x-2)} + \frac{B}{(x-2)^2} + \frac{C}{x+4}

Multiply throughout by (x-2)^2(x+4)

2x2+4=A(x2)(x+4)+B(x+4)+C(x2)22x^2+4 = A(x-2)(x+4) + B(x+4) + C(x-2)^2

We want to find each constant, so get rid of each one by getting values to 0.

Let x = 2
2(2)2+4=A(22)(2+4)+B(2+4)+C(22)22(2)^2 + 4 = A(2-2)(2+4) + B(2+4) + C(2-2)^2
12=6BB=212 = 6B \therefore B = 2

Let x = -4
2(4)2+4=A(42)(4+4)+B(4+4)+C(42)22(-4)^2 + 4 = A(-4-2)(-4+4) + B(-4+4) + C(-4-2)^2
36=36CC=136 = 36C \therefore C = 1

Let x = 0
2(0)2+4=A(02)(0+4)+B(0+4)+C(02)22(0)^2 + 4 = A(0-2)(0+4) + B(0+4) + C(0-2)^2
4=8A+4B+4C4 = -8A + 4B + 4C
Sub in values we know:
4=8A+8+48=8AA=14 = -8A + 8 + 4 \therefore -8 = -8A \therefore A = 1

2x2+4(x2)2(x+4)1(x2)+2(x2)2+1x+4\frac{2x^2+4}{(x-2)^2(x+4)} \equiv \frac{1}{(x-2)} + \frac{2}{(x-2)^2} + \frac{1}{x+4}
Reply 6
Original post by ahyak
(-10x^2+17x-15)/2(2x-1)^2(x+4)

how could I separate the terms? i.e. A/? + B/? etc


You have (-10x^2.....) this is unusual. Have you written the question correctly.
Reply 7
@ L'Evil Fish, thanks for replying, I understand what the method is but what's confusing me is that (2x-1)^2 is also multiplied by 2, and i don't really know how to express that in the form A/? + B/? etc. i've attempted to answer this question many times but still getting the wrong answer unfortunately

@ steve2005, yeah I've checked, it's correct
Original post by ahyak
@ L'Evil Fish, thanks for replying, I understand what the method is but what's confusing me is that (2x-1)^2 is also multiplied by 2, and i don't really know how to express that in the form A/? + B/? etc. i've attempted to answer this question many times but still getting the wrong answer unfortunately

@ steve2005, yeah I've checked, it's correct


You could expand the bracket I guess:

Fraction / (2x+8)(2x-1)^2
Reply 9
Original post by ahyak
(-10x^2+17x-15)/2(2x-1)^2(x+4)

how could I separate the terms? i.e. A/? + B/? etc


10x2+17x152(2x1)2(x+4)=A2x1+B(2x1)2+Cx+4\frac{-10x^2+17x-15}{2(2x-1)^2(x+4)}=\frac{A}{2x-1}+\frac{B}{(2x-1)^2}+\frac{C}{x+4}

You next line should be:

10x2+17x15=A(2)(2x1)(x+4)+B(2)(x+4)+C(2)(2x1)2-10x^2+17x-15=A(2)(2x-1)(x+4)+B(2)(x+4)+C(2)(2x-1)^2
Original post by BabyMaths
10x2+17x152(2x1)2(x+4)=A2x1+B(2x1)2+Cx+4\frac{-10x^2+17x-15}{2(2x-1)^2(x+4)}=\frac{A}{2x-1}+\frac{B}{(2x-1)^2}+\frac{C}{x+4}

You next line should be:

10x2+17x15=A(2)(2x1)(x+4)+B(2)(x+4)+C(2)(2x1)2-10x^2+17x-15=A(2)(2x-1)(x+4)+B(2)(x+4)+C(2)(2x-1)^2


Could you do it this way:

10x2+17x15(2x1)2(2x+8)=A2x1+B(2x1)2+C2x+8\frac{-10x^2+17x-15}{(2x-1)^2(2x+8)}=\frac{A}{2x-1}+\frac{B}{(2x-1)^2}+\frac{C}{2x+8}

Or is that altering it all?

Just wondering.
Reply 11
Original post by L'Evil Fish
Could you do it this way:

10x2+17x15(2x1)2(2x+8)=A2x1+B(2x1)2+C2x+8\frac{-10x^2+17x-15}{(2x-1)^2(2x+8)}=\frac{A}{2x-1}+\frac{B}{(2x-1)^2}+\frac{C}{2x+8}

Or is that altering it all?

Just wondering.


Actually I think that's better than what I said. :smile:
Original post by BabyMaths
Actually I think that's better than what I said. :smile:


Ah okay :smile: I've learnt that we can do it like that then :biggrin:
Reply 13
Original post by L'Evil Fish
Could you do it this way:

10x2+17x15(2x1)2(2x+8)=A2x1+B(2x1)2+C2x+8\frac{-10x^2+17x-15}{(2x-1)^2(2x+8)}=\frac{A}{2x-1}+\frac{B}{(2x-1)^2}+\frac{C}{2x+8}

Or is that altering it all?

Just wondering.


that's not working either.. i'm starting to think that maybe the answer in the textbook is wrong
Original post by ahyak
that's not working either.. i'm starting to think that maybe the answer in the textbook is wrong


I'll post my final answer now.
10x2+17x15(2x1)2(2x+8)=A2x1+B(2x1)2+C2x+8\frac{-10x^2+17x-15}{(2x-1)^2(2x+8)}=\frac{A}{2x-1}+\frac{B}{(2x-1)^2}+\frac{C}{2x+8}

10x2+17x15=A(2x1)(2x+8)+B(2x+8)+C(2x1)2-10x^2+17x-15=A(2x-1)(2x+8) + B(2x+8) + C(2x-1)^2

Let x = 1/2:
-2.5 + 8.5 - 15 = 9B
-9 = 9B
B = -1

then let x = -4:
-160-68-15=81C
-243 = 81C
C = -3

Then 1.. can you continue?

Spoiler

(edited 11 years ago)
Reply 16
12(2x1)1(2x1)232(x+4)\frac{1}{2( 2x-1) }-\frac{1}{{( 2x-1) }^{2}}-\frac{3}{2( x+4) }

Might as well post this..
Original post by BabyMaths
12(2x1)1(2x1)232(x+4)\frac{1}{2( 2x-1) }-\frac{1}{{( 2x-1) }^{2}}-\frac{3}{2( x+4) }

Might as well post this..


:five:
Reply 18
how did you find A...
Original post by BabyMaths
.

:biggrin: you now have 5 gems :wink:
Original post by ahyak
how did you find A...


Choose any value, and sub in the values you already know (in this case B and C) and you'll have one unknown.

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