section 6.3 partial fractions
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Section 6.3 Partial Fractions. Advanced Algebra. What is Partial Fraction Decomposition?. There are times when we are working with Rational Functions of the form when we want to - PowerPoint PPT PresentationTRANSCRIPT
Section 6.3 Partial Fractions
Advanced Algebra
What is Partial Fraction Decomposition?
There are times when we are working with Rational Functions of the form when we want to
split it up into two simpler fractions. The process that we go through to do this is called “partial fraction decomposition”
)()(xgxf
Let’s say you had the following:
14
43
xx
)1)(4(16433
xxxx
)4()4(
)1(4
)1()1(
)4(3
xx
xxx
x
What would you need to do to add them together?
Multiply by a common denominator to both fractions
Write as one fraction and simplify the numerator
)1)(4(137
xx
xThe final answer
Partial Fraction Decomposition is the reverse of what we just did here…
22xx
222
xB
xA
xx
2
22222
xxBx
xxAx
xxxx
BxxA 22
BxAAx 22
A22
A1
BxAx 0
BA0B10
B 1 211
xx
Cancel anything necessary
Multiply by LCD
Break into 2 smaller fractions
Collect like terms and set equal
5 33 1x
x x
3 1A Bx x
Multiply by the common denominator.
5 3 1 3x A x B x
5 3 3x Ax A Bx B Collect like terms together and set them equal to each other.
5x Ax Bx 3 3A B
5 A B 3 3A B Solve two equations with two unknowns.
3235
2 xx
x
5 33 1x
x x
3 1A Bx x
5 3 1 3x A x B x
5 3 3x Ax A Bx B
5x Ax Bx 3 3A B
5 A B 3 3A B Solve two equations with two unknowns.
5 A B 3 3A B
3 3A B
8 4B
2 B 5 2A
3 A
This technique is calledPartial Fractions
12
33
xx
3235
2 xx
x
1310211
2 xx
x
12151310211
2
xB
xA
xxx
)15()12(211 xBxAx
BxBAxAx 1512211
xBxAx 5211 BA 2
BA 5211
BA 5211 BA 224
B77
B1
12 A
A 3
A3
121
153
xx
2
6 72
xx
Repeated roots: we must use two terms for partial fractions.
22 2A Bx x
6 7 2x A x B
6 7 2x Ax A B
6x Ax 7 2A B
6 A 7 2 6 B
7 12 B
5 B
2
6 52 2x x
Sometimes you might get a repeated factor (multiplicity) in the denominator
Partial-Fraction Decomposition Repeated linear factor
2
2
12374
xxxx
22
2
11212374
xC
xB
xA
xxxx
2121374 22 xCxxBxAxx
2212374 222 xCxxBxxAxx
CCxBBxBxAAxAxxx 222374 222
2224 BxAxx CxBxAxx 27 CBA 223
CxBxAxx 27 CBA 223 2224 BxAxx
BA4 CBA 27 CBA 22414
BA 4511
BA 4511
BA 4416
A927 A3
B34B1
C21233
C213
C24
C 2
212
11
23
xxx
Now we have 3 equations with 3 unknowns. Solve like in previous section.
3 2
2
2 4 32 3
x x xx x
If the degree of the numerator is higher than the degree of the denominator, use long division first.
2 3 22 3 2 4 3x x x x x 2x
3 22 4 6x x x 5 3x
2
5 322 3xx
x x
5 323 1xx
x x
3 2 23 1
xx x
(from example one)
What if the denominator has a non-factorable quadratic in it?
21)2)(1(47
22
2
xC
xBAx
xxxx
0242
7 222
CBxBxAx
xCxAx
)1()2)((047 22 xCxBAxxx
Partial Fraction Decomposition can get very complicated, very quickly when there are non-factorable quadratics and repeated linear factors…here is an easy example…
CCxBBxAxAxxx 222 22047
)2)(1(47
2
2
xxxx
Now just solve the 3 by 3 system…
A=3, B=2 and C=4
A nice shortcut if you have non-repeated linear factors—the Heaviside Shortcut—named after mathematician Oliver Heaviside (1850-1925)…
)1)(5(2
xx
x
15)1)(5(2
xB
xA
xxx A
43
1525
B
41
5121
Tell yourself, “if x is 5, then x-5 is 0.” Cover up the x-5 and put 5 in for the x in what is left…
Do the same for the other linear factor
141
543
)1)(5(2
xxxxx Which should probably
be simplified…
22
2 41 1x
x x
irreduciblequadratic
factor
repeated root
22 1 1 1Ax B C Dx x x
first degree numerator
2 2 22 4 1 1 1 1x Ax B x C x x D x
2 3 2 22 4 2 1 1x Ax B x x C x x x Dx D
3 2 2 3 2 22 4 2 2x Ax Ax Ax Bx Bx B Cx Cx Cx C Dx D
A challenging example:
3 2 2 3 2 22 4 2 2x Ax Ax Ax Bx Bx B Cx Cx Cx C Dx D
0 A C 0 2A B C D 2 2A B C 4 B C D
1 0 1 0 02 1 1 1 0
1 2 1 0 20 1 1 1 4
2 r 3
r 1
1 0 1 0 00 3 1 1 40 2 0 0 20 1 1 1 4
2
1 0 1 0 00 1 0 0 10 3 1 1 40 1 1 1 4
3 r 2
r 2
1 0 1 0 00 1 0 0 10 0 1 1 10 0 1 1 3
r 3
1 0 1 0 00 1 0 0 10 3 1 1 40 1 1 1 4
3 r 2
r 2
1 0 1 0 00 1 0 0 10 0 1 1 10 0 1 1 3
r 3
1 0 1 0 00 1 0 0 10 0 1 1 10 0 0 2 2
2
1 0 1 0 00 1 0 0 10 0 1 1 10 0 0 1 1
r 4
1 0 1 0 00 1 0 0 10 0 1 0 20 0 0 1 1
r 3
1 0 0 0 20 1 0 0 10 0 1 0 20 0 0 1 1
1 0 0 0 20 1 0 0 10 0 1 0 20 0 0 1 1
22
2 41 1x
x x
22 1 1 1Ax B C Dx x x
22
2 1 2 11 1 1
xx x x
We can do this problem on the TI-89:
22
2 4expand1 1x
x x
expand ((-2x+4)/((x^2+1)*(x-1)^2))
22 2
2 1 2 11 1 1 1x
x x x x
Of course with the TI-89, we could just integrate and wouldn’t need partial fractions!
3F2