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# partial fraction decomposition

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Could anybody help my decompose this fraction? Thanks!

(x^3-2)/(x^2+x)

Sep 28, 2018

#1
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When you have a hard time, just expand the equation:

(x^3-2)/ (x^2-2) :   (X)(X)(X)-2 / (x)(x)+X

now proceed to eliminate which gives:  (X)-2 / X

X/X= 1 so they cancel out

-2/1 = -2

:)

Sep 28, 2018
#2
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Simplify the following:
(x^3 - 2)/(x^2 + x)
There isn't much you can do with it !!

Factor common terms out of x^2 + x.
Factor x out of x^2 + x:

(x^3 - 2)/(x (x + 1)), or you can write it like this:x^3/(x^2 + x) - 2/(x^2 + x)

Sep 28, 2018
#3
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I am not sure

All I can think to do is division which gave me

$$\frac{x^3-2}{x^2+x}\\ =x-1+\frac{x-2}{x^2+2x} \qquad \text{(This has been edited)}$$

I will leave this to another mathematician.

Continued from earlier

$$\frac{x^3-2}{x^2+x}\\ =x-1+\frac{x-2}{x^2+2x}\\ =x-1+\frac{x-2}{x(x+2)}\\$$

consider just the fraction part

$$\frac{x-2}{x(x+2)}=\frac{A}{x}+\frac{B}{x+2} \qquad \text{for some real A and B}\\ A(x+2)+Bx\equiv x-2\\ Ax+2A+Bx\equiv x-2\\ (A+B)x+2A\equiv x-2\\ so\\ 2A=-2\\ A=-1\\ A+B=1\\ -1+B=1\\ B=2\\ \frac{x-2}{x(x+2)}=\frac{-1}{x}+\frac{2}{x+2}$$

So

$$\frac{x^3-2}{x^2+x} \\ =x-1+\frac{x-2}{x^2+2x} \\ =x-1+\frac{-1}{x}+\frac{2}{x+2}\\ =x-1-\frac{1}{x}+\frac{2}{x+2}\\$$

I think that is right now. What do you think Mathbum and guest?

Sep 28, 2018
edited by Melody  Sep 28, 2018
edited by Melody  Sep 28, 2018
edited by Melody  Sep 29, 2018
#6
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I think the denominator in the second expression is supposed to be x2+x

Guest Sep 29, 2018
#7
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Yes i think you are right and that makes a difference too :/  thanks

Melody  Sep 29, 2018
#8
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I have continued on my original post above   Melody  Sep 29, 2018
#9
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The denominator is x2+x not x2+2x.

Guest Sep 29, 2018
#10
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LOL this must not be my question! it is jinxed Oh well doesn't matter, the technique is the same.

Mathbum can copy the technique and get the original answer correct :)

Melody  Sep 29, 2018
#11
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At the consider just the fraction part Melody, why would you have x-2 in the numerator on the left side?

mathbum  Sep 29, 2018
#13
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Becasue I did the algebraic division first (which I did not display) and x-2 was the remiander.

$$(x^2+x) \text{ divided into } (x^3-2) \;\;\;goes \;\;\; (x-1) \text{ times with a remainder of } (x-2)$$

$$so\\ \frac{x^3-2}{x^2+x}=x-1+\frac{x-2}{x^2+x}$$

Maybe this will help

$$7 \div 2=3\;\;remainder 1\\ so\\ \frac{7}{2}=3+\frac{1}{2}$$

I hope I have not made any typing errors this time. Do you know how to do the algebraic division?

Melody  Sep 29, 2018
edited by Melody  Sep 29, 2018
#4
0 So basically you just have to factor out the common factors on both the denominator or numberstor then cancel them out.

hope this helps!

Sep 28, 2018
#5
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You have not factored this properly Helppls. Still it is good that you had a go.

$$x^2(x-2)=x^3-2x^2$$

Melody  Sep 28, 2018
#12
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(x^3 - 2)  / (x^2 + x)

Partial faction decomposition can only be done whenever the degree of the polynomial in the numerator  is <  the degree of the polynomial in the denominator

Performing polynomial division, we have

x        - 1

x^2 + x  [  x^2   +  0x^2 + 0x   - 2 ]

x^3   + x^2

___________________

-x^2  + 0x

-x^2  - 1x

_________

x    -  2

The result  is   ( x - 1)  +  ( x - 2) / ( x^2 + x)

We can decompose the  remainder  as

( x - 2) / [ x ( x + 1)   =   A / x   +   B / ( x + 1)

Multiply through by    x ( x + 1)

x -  2  =  A(x + 1)  + Bx   simplify

x - 2 =  (A + B)x  + A       equate coefficients

A + B  = 1

A  =  -2

Which means that B  = 3

So...the complete  decomposition is

(x -1 )  - [ 2 / x ]   +  [ 3 / ( x + 1) ]   Sep 29, 2018