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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 factors𝑥+3
𝑥2+3 𝑥+2≡
𝑥+3(𝑥+2 ) (𝑥+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 factors𝑥+3
𝑥2+3 𝑥+2≡
𝑥+3(𝑥+2 ) (𝑥+1 )
𝐴𝑥+2
+𝐵𝑥+1
Next
Repeated factors
Quadratic factors
Linear factors
Introduction
Linear factors𝑥+3
𝑥2+3 𝑥+2≡
𝑥+3(𝑥+2 ) (𝑥+1 )
𝐴𝑥+2
+𝐵𝑥+1
𝐴 (𝑥+1 )+𝐵(𝑥+2)𝑥2+3 𝑥+2
and are found by putting the expression in this form…
Next
Repeated factors
Quadratic factors
Linear factors
Introduction
Linear factors𝑥+3
𝑥2+3 𝑥+2≡
𝑥+3(𝑥+2 ) (𝑥+1 )
𝐴𝑥+2
+𝐵𝑥+1
𝐴 (𝑥+1 )+𝐵(𝑥+2)𝑥2+3 𝑥+2
𝐴 (𝑥+1 )+𝐵 (𝑥+2 )≡𝑥+3
𝑥+3𝑥2+3 𝑥+2
≡−1𝑥+2
+2
𝑥+1
… and solving this equivalence
Next
Repeated factors
Quadratic factors
Linear factors
Introduction
Cover-up method
𝑓 (𝑥 )= 𝑥(𝑥−2 ) (𝑥−3 )
≡𝐴
𝑥−2+
𝐵𝑥−3
𝐴= 𝑓 (2 )= 2(𝑥−2 ) (2−3 )
=−2
𝐵= 𝑓 (3 )= 3(3−2 ) (𝑥−3 )
=3
𝑥(𝑥−2 ) (𝑥−3 )
≡−2𝑥−2
+3
𝑥−3
If we “cover-up” the factor associated with the value we want to fin and then evaluate at it’s 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 factors𝑥2+1
(𝑥2+2)(𝑥−1)
𝐴𝑥+𝐵𝑥2+2
+𝐶𝑥−1
Next
Repeated factors
Quadratic factors
Linear factors
Introduction
If there is a quadratic factor we have partial fractions of this form.
Higher order factors𝑥2+1
(𝑥2+2)(𝑥−1)
𝐴𝑥+𝐵𝑥2+2
+𝐶𝑥−1
(𝐴𝑥+𝐵) (𝑥−1 )+𝐶 (𝑥2+2)(𝑥2+2)(𝑥−1)
(𝐴𝑥+𝐵) (𝑥−1 )+𝐶 (𝑥2+2)≡𝑥2+1Next
Repeated factors
Quadratic factors
Linear factors
Introduction
Rearrange to gain this equivalence.
Higher order factors𝑥2+1
(𝑥2+2)(𝑥−1)
𝐴𝑥+𝐵𝑥2+2
+𝐶𝑥−1
(𝐴𝑥+𝐵) (𝑥−1 )+𝐶 (𝑥2+2)(𝑥2+2)(𝑥−1)
(𝐴𝑥+𝐵) (𝑥−1 )+𝐶 (𝑥2+2)≡𝑥2+1
𝐴=13,𝐵=
13,𝐶=
23
𝑥2+1(𝑥2+2)(𝑥−1)
≡𝑥+1
3 (𝑥2+2 )+ 23 (𝑥−1 )Next
Repeated factors
Quadratic factors
Linear factors
Introduction
We can no compare coefficients to find our missing values of and .
Repeated factors𝑥−1
(𝑥+1) (𝑥−2 )2
−29
𝑥+1+
𝐵𝑥−2
+𝐶
(𝑥−2 )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 factors𝑥−1
(𝑥+1) (𝑥−2 )2
−29
𝑥+1+
𝐵𝑥−2
+𝐶
(𝑥−2 )2
𝑥−1≡(− 29 ) (𝑥−2 )2+𝐵 (𝑥+1 ) (𝑥−2 )+𝐶 (𝑥+1 )
𝑥=2 ⟹ 𝐶=13 Next
Repeated factors
Quadratic factors
Linear factors
Introduction
Set to eliminate
Repeated factors𝑥−1
(𝑥+1) (𝑥−2 )2
−29
𝑥+1+
𝐵𝑥−2
+𝐶
(𝑥−2 )2
𝑥−1≡(− 29 ) (𝑥−2 )2+𝐵 (𝑥+1 ) (𝑥−2 )+𝐶 (𝑥+1 )
𝑥=2 ⟹ 𝐶=13
0=−29+𝐵⇒𝐵=
29
𝑥−1(𝑥+1) (𝑥−2 )2
≡−2
9(𝑥+1)+
29 (𝑥−2 )
+1
3 (𝑥−2 )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:
𝐴𝑎𝑥+𝑏
𝐴𝑥+𝐵𝑎𝑥2+𝑏𝑥+𝑐
𝐴𝑎𝑥+𝑏
+𝐵
(𝑎𝑥+𝑏)2
Next
Repeated factors
Quadratic factors
Linear factors
Introduction