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Partial fraction decomposition of the rational expression 
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#1
Apr610, 03:25 PM

P: 16

1. The problem statement, all variables and given/known data
Write the partial fraction decomposition of the rational expression. Check your result algebraically. (x^{2} – 7x + 16)/[(x + 2)(x2 – 4x + 5)] 3. The attempt at a solution [A/(x+2)] + [(Bx+C)/(x^{2}4x+5)] x^{2}7x+16= A(X^{2}4x+5)+(Bx+C)(x+2) X^{2}7x+16=[A+B]x^{2}+[4A+B+C]x+[5A+C] A+B=1 => B=1A 4A+B+C=7 => Will need to plugin later 5A+C=16 => C=165A 4A+B+C=7 => 4A+1A+165A= 10A=17 A=(17/10) Now I plug A back into the others B=1(17/10) => B=(27/10) C=165(17/10) => C= (49/2) Result: [(17/10)/(x+2)] + [((27x/10)+(49/2))/(x^{2}4x+5)] However, I am being told the answer is wrong. What is it that I am doing wrong? 


#2
Apr610, 03:35 PM

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#3
Apr610, 03:47 PM

P: 16

I realize there is a mistake, but you could please elaborate upon my mistake without giving me the answer? Your response did not make it obvious or clear for me. Maybe a more indepth response would help me better understand my own mistake. Thank you in advance for taking the time to help me with this problem!
Is it suppose to be X^{2}7x+16=[A+B]x^{2}+[4A+2B+C]x+[5A+2C] Is that correct? 


#4
Apr610, 04:27 PM

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Partial fraction decomposition of the rational expression
That's what I get



#5
Apr610, 04:45 PM

P: 16

A+B=1 => A=1+B 4A+2B+C=7 => 5A+2C=16 => 


#6
Apr610, 05:19 PM

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#7
Apr610, 05:30 PM

P: 16




#8
Apr610, 05:49 PM

P: 163

Use one of the three equations to get a substitution formula. For example, you have A=1B so far. By substituting into the other two equations, you should be able to reduce your system down to 2 equations and 2 unknowns. Go from there...



#9
Apr610, 08:06 PM

P: 180

[QUOTE=mistalopez;2658953]1. The problem statement, all variables and given/known data
Write the partial fraction decomposition of the rational expression. Check your result algebraically. (x^{2} – 7x + 16)/[(x + 2)(x2 – 4x + 5)] 3. The attempt at a solution [A/(x+2)] + [(Bx+C)/(x^{2}4x+5)] You need to change the Bx+C. That polynomial can be reduced. So change it to a B and a C. 


#10
Apr610, 09:29 PM

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[QUOTE=GreatEscapist;2659476]



#11
Apr610, 09:38 PM

P: 180

[QUOTE=Mark44;2659665]



#12
Apr710, 01:00 AM

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P: 21,216

Yes, I'm sure. x^{2} 4x + 5 can't be factored into linear factors with real coefficients.



#13
Apr710, 06:09 PM

P: 180

Oh my god, how stupid of me. I read that wrong.
I'm so sorry. >.< 


#14
Apr710, 06:20 PM

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P: 21,216




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