Partial fraction decomposition of the rational expression


by mistalopez
Tags: decomposition, expression, fraction, partial, rational
mistalopez
mistalopez is offline
#1
Apr6-10, 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.

(x2 7x + 16)/[(x + 2)(x2 4x + 5)]

3. The attempt at a solution

[A/(x+2)] + [(Bx+C)/(x2-4x+5)]

x2-7x+16= A(X2-4x+5)+(Bx+C)(x+2)
X2-7x+16=[A+B]x2+[-4A+B+C]x+[5A+C]

A+B=1 => B=1-A
-4A+B+C=-7 => Will need to plugin later
5A+C=16 => C=16-5A

-4A+B+C=-7 => 4A+1-A+16-5A= -10A=17 A=(-17/10)

Now I plug A back into the others

B=1-(-17/10) => B=(27/10)

C=16-5(-17/10) => C= (49/2)


Result: [(-17/10)/(x+2)] + [((27x/10)+(49/2))/(x2-4x+5)]

However, I am being told the answer is wrong. What is it that I am doing wrong?
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Mark44
Mark44 is offline
#2
Apr6-10, 03:35 PM
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P: 21,081
Quote Quote by mistalopez View Post
1. The problem statement, all variables and given/known data

Write the partial fraction decomposition of the rational expression. Check your result algebraically.

(x2 7x + 16)/[(x + 2)(x2 4x + 5)]

3. The attempt at a solution

[A/(x+2)] + [(Bx+C)/(x2-4x+5)]

x2-7x+16= A(X2-4x+5)+(Bx+C)(x+2)
X2-7x+16=[A+B]x2+[-4A+B+C]x+[5A+C]
I see a couple of mistakes on the line above, in the coefficients of x and the constant term.
Quote Quote by mistalopez View Post

A+B=1 => B=1-A
-4A+B+C=-7 => Will need to plugin later
5A+C=16 => C=16-5A

-4A+B+C=-7 => 4A+1-A+16-5A= -10A=17 A=(-17/10)

Now I plug A back into the others

B=1-(-17/10) => B=(27/10)

C=16-5(-17/10) => C= (49/2)


Result: [(-17/10)/(x+2)] + [((27x/10)+(49/2))/(x2-4x+5)]

However, I am being told the answer is wrong. What is it that I am doing wrong?
mistalopez
mistalopez is offline
#3
Apr6-10, 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 X2-7x+16=[A+B]x2+[-4A+2B+C]x+[5A+2C]

Is that correct?

Mark44
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#4
Apr6-10, 04:27 PM
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Partial fraction decomposition of the rational expression


That's what I get
mistalopez
mistalopez is offline
#5
Apr6-10, 04:45 PM
P: 16
Quote Quote by Mark44 View Post
That's what I get
But now I am stuck because I can only get one value.

A+B=1 => A=1+B

-4A+2B+C=-7 =>

5A+2C=16 =>
Mark44
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#6
Apr6-10, 05:19 PM
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Quote Quote by mistalopez View Post
But now I am stuck because I can only get one value.

A+B=1 => A=1+B
Mistake above. A = 1 - B
Quote Quote by mistalopez View Post

-4A+2B+C=-7 =>

5A+2C=16 =>
You have three equations in three unknowns, so you should be able to solve for them.
mistalopez
mistalopez is offline
#7
Apr6-10, 05:30 PM
P: 16
Quote Quote by Mark44 View Post
Mistake above. A = 1 - B


You have three equations in three unknowns, so you should be able to solve for them.
That was my question. How am I suppose to solve for 3 unknowns when I cannot cancel anything out. I need atleast 2 known variables to solve any of the equations. Do I subtract equations or some sort of method?
rs1n
rs1n is offline
#8
Apr6-10, 05:49 PM
P: 163
Use one of the three equations to get a substitution formula. For example, you have A=1-B 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...
GreatEscapist
GreatEscapist is offline
#9
Apr6-10, 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.

(x2 7x + 16)/[(x + 2)(x2 4x + 5)]

3. The attempt at a solution

[A/(x+2)] + [(Bx+C)/(x2-4x+5)]

You need to change the Bx+C. That polynomial can be reduced. So change it to a B and a C.
Mark44
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#10
Apr6-10, 09:29 PM
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P: 21,081
[QUOTE=GreatEscapist;2659476]
Quote Quote by mistalopez View Post
1. The problem statement, all variables and given/known data

Write the partial fraction decomposition of the rational expression. Check your result algebraically.

(x2 7x + 16)/[(x + 2)(x2 4x + 5)]

3. The attempt at a solution

[A/(x+2)] + [(Bx+C)/(x2-4x+5)]

You need to change the Bx+C. That polynomial can be reduced. So change it to a B and a C.
NO! mistalopez has the correct decomposition.
GreatEscapist
GreatEscapist is offline
#11
Apr6-10, 09:38 PM
P: 180
[QUOTE=Mark44;2659665]
Quote Quote by GreatEscapist View Post
NO! mistalopez has the correct decomposition.
Are you sure? We learned to break down polynomials to factors first, before making it a Bx+c. We learned that the only time you don't use that is when it does not come out evenly in a factor.
Mark44
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#12
Apr7-10, 01:00 AM
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P: 21,081
Yes, I'm sure. x2 -4x + 5 can't be factored into linear factors with real coefficients.
GreatEscapist
GreatEscapist is offline
#13
Apr7-10, 06:09 PM
P: 180
Oh my god, how stupid of me. I read that wrong.
I'm so sorry. >.<
Mark44
Mark44 is offline
#14
Apr7-10, 06:20 PM
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P: 21,081
Quote Quote by GreatEscapist View Post
Oh my god, how stupid of me. I read that wrong.
I'm so sorry. >.<
We all screw up from time to time...


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