7-4a partial fraction decomposition
TRANSCRIPT
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8/8/2019 7-4A Partial Fraction Decomposition
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Warm UpGraph
1
8)(
2
3
x
xxg
noneAV ..
xyAS ..
)0,2)(8,0(
2
2
2
2
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Partial Fraction
DecompositionChapter 2.7
Day 1 of 3
Tuesday, September 17, 2013
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Partial Fraction DecompositionThe purpose of PFD is to separate a fraction into
a sum of pre LCD terms.
)2)(4(
165
xx
x
2
3
4
2
xx
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Partial Fraction Decomposition1) Check if the fraction is proper (The power
in the numerator is less than that of the
denominator.) If it is not then divide. 6
72
xx
x
2) Factor the denominator.
3) Set up the factors. Dont
forget any special rules.
(Details to come.)
)2)(3(
7
xx
x
23)2)(3(7
xB
xA
xxx
4) Solve for A, B, etc. Use matching or
substitution method.
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Substitution Method4.1A) Multiply both sides by the LCD
2)2)(3(
3)2)(3(
)2)(3(
7)2)(3(
x
Bxx
x
Axx
xx
xxx
BxAxx )3()2(7
4.2A) Substitute a value for x that will eliminate A.
BA )32()22(72 2x
BA 505
B55
B1
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Substitution Method4.3A) Substitute a value for x that will eliminate B.
BA )33()23(73 3x
BA 0510
A2
5) Substitute A, B, etc. back into the original expansion.
23
x
B
x
A
2
1
3
2
xx
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Improper Fractions
xxxxxxx
23
234
262062
1) Check if the fraction is proper (The power in the numerator is
less than that of the denominator.) If it is not then divide.
62062 234 xxxxxxx 23 2
x
234 2 xxx ( )
6205 2 xx
xxx
xxx
23
2
2
62051A) Perform the PFD on theremainder over the divisor.
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Repeated Factor Rule2) Factor the denominator.
xxx
xx
23
2
2
6205
)12(
62052
2
xxx
xx2
2
)1(
6205
xx
xx
3) Set up the factors. Dont forget any special rules.
3A) The repeated factor rule. If a factor occurs more than
once then a term for each occurrence starting at power1 and up to power n must be placed.
22
2
)1(1)1(
6205
x
C
x
B
x
A
xx
xx
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Partial Fraction Decomposition4) Solve for A, B and C.
22
2
)1(1)1(
6205
x
C
x
B
x
A
xx
xx
4.1A) Multiply both sides by the LCD
2
222
2
22
)1()1(
1)1()1(
)1(
6205)1(
x
Cxx
x
Bxx
x
Axx
xx
xxxx
CBA )1()11)(1()11(6)1(20)1(5 22
4.2A) Substitute a value for x that will eliminate A.
xCBxxAxxx )1()1(6205 22
1x
CBA 006205
C9 C9
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Partial Fraction Decomposition4.3A) Substitute a value for x that will eliminate B.
CBA 0)10(0)10(6)0(20)0(5 22 0x
A6
4.4A) Substitute back A and C
xCBxxAxxx )1()1(6205 22
9)1(6)1(6205 22 xBxxxxx
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Partial Fraction Decomposition
4.5A) Substitute in any value for x but 0 and -1.
1x 9)1()11(16)11(6)1(20)1(522
B
92246205 B922431 B
B23331
B22
B1
5) Substitute A, B and C back into the original expansion.
2)1(
9
1
16
xxxx
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Partial Fraction Decomposition
2)2(
43
x
x2)2(2
x
B
x
A
BAxx )2(43
BA )22(4)2(32x
BA )0(46
B2
2)2(43 Axx
0x 2)20(4)0(3 A
224 A
A26
A3
2)2(
2
2
3
xx
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Check Your Work
2)2(
2
2
3
xx
22 )2(
2
)2(
)2(3
xx
x
22 )2(
2
)2(
63
xx
x
2)2(
263
x
x
2)2(
43
x
x