Download Document

PAR
TIAL
+
FRAC TION
DECOMPOSITION
Let’s add the two fractions below. We need a common
denominator:
 x2


 x2
4
3  x 3



x

3


x3 x2
4 x  8  3x  9
x  2x  3
7x 1
2
x  x6
In this section we are going to learn
how to take this answer and
“decompose” it meaning break
down into the fractions that were
added together to get this answer
7x 1
2
x  x6
We start by factoring the denominator.
There could have been a fraction for
each factor of the denominator but we
don’t know the numerators so we’ll
call them A and B.
A
B
7x 1


x  3x  2 x  3 x  2
(x+3)(x-2)
7 x  1  Ax  2  Bx  3
Now we’ll clear this equation of fractions by multiplying
every term by the common denominator.
7x 1
2
x  x6
“The Convenient x Method for Solving”
This equation needs to be true for any
value of x.
7 x  1  Ax  2  Bx  3
Let x = -3
B(0) = 0
7 3  1  A 3  2  B 3  3
 20  5 A
A4
We pick an x that will “conveniently” get rid of one of the
variables and solve for the other.
7x 1
2
x  x6
Now we’ll “conveniently” choose x to
be 2 to get rid of A and find B.
7 x  1  Ax  2  Bx  3
Let x = 2
A(0) = 0
72  1  A2  2  B2  3
A4
15  5B
B3
3
7x 1
A4
B


x  3x  2 x  3 x  2
Summary of Partial Fraction Decomposition
When Denominator Factors Into Linear
Factors (Factors of first degree)
Next we’ll look at repeated factors and quadratic factors
Use “convenient” x method to find A, B, etc.
Clear equation of fractions
Set fraction equal to sum of fractions with
each factor as a denominator using A, B,
etc. for numerators
Factor the denominator
Partial Fraction Decomposition With
Repeated Linear Factors
When the denominator has a repeated linear factor, you
need a fraction with a denominator for each power of the
factor.
A
B
C
x 2



2
2
x  1x  2 x  1 x  2 x  2
2
x
2

 2  Ax  2  Bx  1x  2  C x  1
1
2

 2  A1  2  B1  11  2  C 1  1
1
3  9A A 
3
Let x = 1
2
2
1
3
A
2
B3
x 2
-2
C


2 
x  1x  2 x  1 x  2 x  22
2
x
2
(2)

 2  Ax  2  Bx  1x  2  C x  1
2
2

 2  A 2  2  B 2  1 2  2  C  2  1
2
6  3C C  2
Let x = -2
To find B we put A and C in and choose x to be any other
number. Let x = 0


1
2
0  2  0  2  B0  10  2  20  1
3
2
4
2   2B  2
3
4
2B 
3
4 2
B 
6 3
Partial Fraction Decomposition With
Quadratic Factors
When the denominator has a quadratic factor (that won’t
factor), you need a fraction with a linear numerator.
1
x  1 x 2  4




A
Bx  C

 2
x 1 x  4
1  A x 2  4  Bx  C x  1
The convenient x method doesn’t work as nicely on
these kind so we’ll use the “equating coefficients”
method. First multiply everything out.
1
x  1 x 2  4





1
5
A
x 1
1
1

5Bx  C
5

x 4
2
1  A x  4  Bx  C x  1
2
1
A
5
1
C
5
1  Ax  4 A  Bx  Bx  Cx  C
2
Look at x2 terms:
2
0=A+B A=-B
0=B+C C=-B
1
B
Look at x terms:
5
No x2 terms on left
Look at terms
1 = 4A + C 1 = 4(-B) + (-B)
terms on left
withNo
noxx’s:
Solve these. Substitution would probably be easiest.
Look at the kinds of terms on each side and equate
coefficients (meaning put the coefficients = to each other)
Partial Fraction Decomposition With
Repeated Quadratic Factors
When the denominator has a repeated quadratic factor
(that won’t factor), you need a fraction with a linear
numerator for each power.
x x
x
3
2
2

4
2
Ax  B Cx  D
 2

2
2
x  4 x  4


x  x   Ax  B  x  4  Cx  D
3
2
2
multiply out
x  x  Ax  4 Ax  Bx  4 B  Cx  D
3
2
3
2
equate coefficients of various kinds of terms (next screen)
x x
x
3
2
2

4
2
1 -4
1 B
Ax
Cx -4D
 2

2
2
x  4 x  4
1 x 3  1 x 2  Ax3  4 Ax  Bx 2  4 B  Cx  D
Look at x3 terms:
1=A
Look at x2 terms:
1=B
Look at x terms:
0 = 4A+C
0 = 4(1)+C
Look at terms
with no x:
0 = 4B+D
0 = 4(1)+D -4 = D
-4 =C