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Perplex
Perplex
  • Dashboard
Topics
Exponents & LogarithmsRounding & ErrorSequences & SeriesCounting & BinomialsProof and ReasoningComplex NumbersAlgebra Skills
Cartesian plane & linesQuadraticsFunction TheoryTransformations & asymptotesPolynomials
2D & 3D GeometryTrig equations & identitiesVectors
ProbabilityDescriptive StatisticsBivariate StatisticsDistributions & Random Variables
DifferentiationIntegrationDifferential EquationsMaclaurin
Paper 3
Plus
Calculator Skills
Review VideosFormula BookletAll Study Sets
BlogLanding Page
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Perplex

Algebra Skills (Lesson 1/2)

Partial fractions

1 / 3

Linear partial fractions

When the denominator of a fraction can be factored, it is possible to split it into the sum of ​2​ pieces:

​
x2+2x−34x−8​≡(x+3)(x−1)4x−8​
​

(where ​≡​ means "equal for all ​x​")

Remember that when we add two fractions, we cross multiply the denominators:

​
x+3A​+x−1B​≡(x+3)(x−1)A(x−1)+B(x+3)​
​

Then we can write

​
4x−8≡A(x−1)+B(x+3)
​


The easiest way to solve this is to chose a value of ​x​ that eliminates ​A​ or ​B, eg ​x=1:

​
4⋅1−8=A⋅(1−1)+B(1+3)
​
​
−4=4B⇒B=−1
​

and therefore

​
4x−8=Ax−A−x−3⇒A=5
​

Algebra Skills (Lesson 1/2)

Partial fractions

1 / 3

Linear partial fractions

When the denominator of a fraction can be factored, it is possible to split it into the sum of ​2​ pieces:

​
x2+2x−34x−8​≡(x+3)(x−1)4x−8​
​

(where ​≡​ means "equal for all ​x​")

Remember that when we add two fractions, we cross multiply the denominators:

​
x+3A​+x−1B​≡(x+3)(x−1)A(x−1)+B(x+3)​
​

Then we can write

​
4x−8≡A(x−1)+B(x+3)
​


The easiest way to solve this is to chose a value of ​x​ that eliminates ​A​ or ​B, eg ​x=1:

​
4⋅1−8=A⋅(1−1)+B(1+3)
​
​
−4=4B⇒B=−1
​

and therefore

​
4x−8=Ax−A−x−3⇒A=5
​