Partial Fraction Decomposition
Many times,it is possibleto split a complicatedrationalexpressioninto a sumof smaller,simpler
expressions.
This sumis calledthepartialfraction decomposition
of theoriginalrationalexpression.
Settingup thedecompositiondependson the factorsof the denominator,but.theprocessof solvingthe
decomposition
is the same.
In general,thedecomposition
will haveonefractionfor eachfactorin the denominator.If the factoris
linear, the numeratorof the fractionis just anunknownconstant,like A. If the factoris quadractic,the
numeratorwill havea linearform, like A"r+B.
To solvea decomposition,
we solvefor theunknownnumerators.Let's look at a simpleexample.
Ex. Find the partial fractiondecomposition
of
^
*I)o
.( x - 4 ) ( x + 2
^,) .
Stepl: Setup the partialfractiondecomposition
with the unknownconstantsA, B, C, etc,in the
numeratorof thedecomposition.
*t!)a
A.* B=,andourgoalistofindAandB.
,',
=,=
(x-4)(x+2) x-4 x+2'
Step2: Multiply both sidesof the resultingequationby the leastcommondenominator.
.'.(x4)(x+zt
Uffi=(x-
4)(x+zx**4t
Step3: Simplify both sidesof the equation.Everydenominatorshouldcancelout.
:.@24)@.d)
dh
=(r,-4)(x
. r)
A+
(x- 4)O
Ft)
&
: . x + 1 4- ( x + 2 ) A + @ - g B
Step4: Write both sidesin descending
powers,equatecoefficientsof the like terms.Usedistributivelaw
on our equationfrom step3, andthengrouplike termtogether.
x + 1 4= ( A + B ) x + ( 2 A - 4 8 ) ,
[l=A+B
"'1to=2A.6
4B
Step5: Solvefor the resultinglinearsystemfor e, B, C, etc.
Usingsubstitution,we get A = 3 andB = -2. (Checkthis yourself'!)
Step6: Substitutethe valuefor A andB into stepone
A*B_3_2
...x+t4(x-4)(x+2) x-4 x+2 x-4 x+2
We now haveour partial fractiondecomposition!Our original factionhasbeensplit into two smaller
fractions.
Thereis an alternatemethodwe couldusein placeof steps4 and5. We cansubstitutevaluesofx that
makeunknowncoefficientsdisappear.
In our example,
-2
Steps4 and5(altemate):Plug in for x, in orderto makethe term (x + 2)A disappear.This givesus
.'.(-2) +r4 = (-2 + 2)A + (-2 - 4)B
.' .12= - 68
:.8=-2
Partial FractionDecomposition
Similarly, we can plug in 4 for x to make the term @ - $B disappear.Solving the resulting equation
gives us A:3, just like in our original method.
Many people find this alternatemethod easier,but unfortunately, it does not always work.
Let's look at a larger, more complicated example. This example has arepeated linear factor and a
quadratic factor:
xt - x' + 4x -l 4
.UX]-:----:The denominatorhas a linear factor, x, repeatedtwice, and a quadratic
x"(x'+I)
factor,xz +1.
Stepl: Setup the partialfractiondecomposition
with the unknownconstantsA, B, C, etc,in the
numeratorof the decomposition.
xt - x' +4x-l
A B C x +D
w|)=-,*7*T|4Sincethelinearfactorisrepeatedtwice,itneedstwo
fractions,onewith an exponentof l, anotherwith an exponentof 2.
Step2: Multiplyboth sidesof the resultingequationby the leastcommondenominator.
(*')(r' * r)' *' :r' l l:r-r = (x2)(x2
+D4 + (x2)7x2
+ D' 4 + 62y(x2
*g- ' 9!!2
x'(x'+l)
x
xz
x 2+ l
Step3: Simplifuboth sidesof the equation.Everydenominatorshouldcancelout.
xt - x' + 4x -l = (x)(xz+l)A+ (xz+l)B + (x'z)(Cx
+ D)
Step4: Write both sidesin descending
powers,equatecoefficientsof the like terms.We will haveto use
distributivelaw to simplify our equationform step3, andthengoup like termstogether.
x t -x'+4 x-I= xt(A +C )+ x' 18 + D) + x( A) +B ( Checkthis
for your selfl)
lr=A+c
l-1= B+D
"'70=
n
[-r=r
Step5: Solvefor the resultinglinearsystemfor A, B, C, etc.
Using substitution,we find that A=4, B:-1, C:-3, andD=0. (Checkthis for yourself!)
Step6: Substitutethe valuefor the constantsinto stepone.
xt-x'+4x-I
A B Cx+D 4 -l -3x+0
=-I-f-=_t_I_
x'1x' +I1
x
x'
xt +1
x
x'
xz+l
We now haveour partialfractiondecomposition!In many situations,it will be easierto work with these
simplerfractionsthanour original,complicatedfraction.
Unfortunately,this is an examplewhereour alternatemethodwill not get us very far. For this reason,it is
importantto know morethanonemethodof solvingpartial fractiondecompositions.