www.le.ac.uk partial fractions department of mathematics university of leicester
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www.le.ac.uk
Partial Fractions
Department of MathematicsUniversity of Leicester
Content
Quadratic factors
Linear factors
Repeated factors
Introduction
Introduction
Fractions whose algebraic sum is a given fraction are called partial fractions.
E.g.
and are partial fractions of since
Next
Repeated factors
Quadratic factors
Linear factors
Introduction
Linear factorsπ₯+3
π₯2+3 π₯+2β‘
π₯+3(π₯+2 ) (π₯+1 )
The decomposition of a given fraction into partial fractions is achieved by first factorising the denominator
Next
Repeated factors
Quadratic factors
Linear factors
Introduction
If the factors are linear then we will have partial fractions of this form
Linear factorsπ₯+3
π₯2+3 π₯+2β‘
π₯+3(π₯+2 ) (π₯+1 )
π΄π₯+2
+π΅π₯+1
Next
Repeated factors
Quadratic factors
Linear factors
Introduction
Linear factorsπ₯+3
π₯2+3 π₯+2β‘
π₯+3(π₯+2 ) (π₯+1 )
π΄π₯+2
+π΅π₯+1
π΄ (π₯+1 )+π΅(π₯+2)π₯2+3 π₯+2
and are found by putting the expression in this formβ¦
Next
Repeated factors
Quadratic factors
Linear factors
Introduction
Linear factorsπ₯+3
π₯2+3 π₯+2β‘
π₯+3(π₯+2 ) (π₯+1 )
π΄π₯+2
+π΅π₯+1
π΄ (π₯+1 )+π΅(π₯+2)π₯2+3 π₯+2
π΄ (π₯+1 )+π΅ (π₯+2 )β‘π₯+3
π₯+3π₯2+3 π₯+2
β‘β1π₯+2
+2
π₯+1
β¦ and solving this equivalence
Next
Repeated factors
Quadratic factors
Linear factors
Introduction
Cover-up method
π (π₯ )= π₯(π₯β2 ) (π₯β3 )
β‘π΄
π₯β2+
π΅π₯β3
π΄= π (2 )= 2(π₯β2 ) (2β3 )
=β2
π΅= π (3 )= 3(3β2 ) (π₯β3 )
=3
π₯(π₯β2 ) (π₯β3 )
β‘β2π₯β2
+3
π₯β3
If we βcover-upβ the factor associated with the value we want to fin and then evaluate at itβs zero we will achieve the value we were looking for.
You can check this is correct with the previous method.
This only works with linear factors!
Next
Repeated factors
Quadratic factors
Linear factors
Introduction
Higher order factorsπ₯2+1
(π₯2+2)(π₯β1)
π΄π₯+π΅π₯2+2
+πΆπ₯β1
Next
Repeated factors
Quadratic factors
Linear factors
Introduction
If there is a quadratic factor we have partial fractions of this form.
Higher order factorsπ₯2+1
(π₯2+2)(π₯β1)
π΄π₯+π΅π₯2+2
+πΆπ₯β1
(π΄π₯+π΅) (π₯β1 )+πΆ (π₯2+2)(π₯2+2)(π₯β1)
(π΄π₯+π΅) (π₯β1 )+πΆ (π₯2+2)β‘π₯2+1Next
Repeated factors
Quadratic factors
Linear factors
Introduction
Rearrange to gain this equivalence.
Higher order factorsπ₯2+1
(π₯2+2)(π₯β1)
π΄π₯+π΅π₯2+2
+πΆπ₯β1
(π΄π₯+π΅) (π₯β1 )+πΆ (π₯2+2)(π₯2+2)(π₯β1)
(π΄π₯+π΅) (π₯β1 )+πΆ (π₯2+2)β‘π₯2+1
π΄=13,π΅=
13,πΆ=
23
π₯2+1(π₯2+2)(π₯β1)
β‘π₯+1
3 (π₯2+2 )+ 23 (π₯β1 )Next
Repeated factors
Quadratic factors
Linear factors
Introduction
We can no compare coefficients to find our missing values of and .
Repeated factorsπ₯β1
(π₯+1) (π₯β2 )2
β29
π₯+1+
π΅π₯β2
+πΆ
(π₯β2 )2
Next
Repeated factors
Quadratic factors
Linear factors
Introduction
If there is a repeated factor we have partial fractions of this form. The numerator of the linear factor was found using the cover-up method.
Repeated factorsπ₯β1
(π₯+1) (π₯β2 )2
β29
π₯+1+
π΅π₯β2
+πΆ
(π₯β2 )2
π₯β1β‘(β 29 ) (π₯β2 )2+π΅ (π₯+1 ) (π₯β2 )+πΆ (π₯+1 )
π₯=2 βΉ πΆ=13 Next
Repeated factors
Quadratic factors
Linear factors
Introduction
Set to eliminate
Repeated factorsπ₯β1
(π₯+1) (π₯β2 )2
β29
π₯+1+
π΅π₯β2
+πΆ
(π₯β2 )2
π₯β1β‘(β 29 ) (π₯β2 )2+π΅ (π₯+1 ) (π₯β2 )+πΆ (π₯+1 )
π₯=2 βΉ πΆ=13
0=β29+π΅βπ΅=
29
π₯β1(π₯+1) (π₯β2 )2
β‘β2
9(π₯+1)+
29 (π₯β2 )
+1
3 (π₯β2 )2
Next
Repeated factors
Quadratic factors
Linear factors
Introduction
Comparing the coefficients of we can find .
Summary
A linear factor gives a partial fraction of the form:
A quadratic factor gives a partial fraction of the form:
A repeated factor gives a partial fraction of the form:
π΄ππ₯+π
π΄π₯+π΅ππ₯2+ππ₯+π
π΄ππ₯+π
+π΅
(ππ₯+π)2
Next
Repeated factors
Quadratic factors
Linear factors
Introduction
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