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Perplex
Perplex
  • Dashboard
Topics
Exponents & LogarithmsRounding & ErrorSequences & SeriesCounting & BinomialsProof and ReasoningComplex NumbersAlgebra Skills
Paper 3
Plus
Calculator Skills
Review VideosFormula BookletAll Study Sets
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Perplex
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Algebra Skills
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Partial fractions
3 by 3 Systems of equations
Partial fractions
Algebra Skills

Partial fractions

0 of 0 exercises completed

Decomposing rational expressions into partial fractions, including linear factors such as ​x+3A​+x−1B​​ and more complex forms, by cross-multiplying and solving for the unknown constants.

Linear partial fractions
AHL AA 1.11

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
​
Non-linear partial fractions
AHL AA 1.11

Aside from expressions of the form ​(x−c)(x−d)ax+b​, there are all sorts of types of partial fractions. For more complicated cases, exam questions will specify the form of the partial fractions decomposition, and you simply need to solve for the variables.


Solve for ​A​ and ​B​ in the equation ​x5−2x4+x−22x4+x​≡x4+1A​+x−2B​​


Cross multiplying:

​
x4+1A​+x−2B​=x5−2x4+x−2A(x−2)+B(x4+1)​
​

So

​
A(x−2)+B(x4+1)=2x4+x
​

We notice immediately that ​B=2, so

​
Ax−2A+2x4+2=2x4+x⇒A=1
​

Nice work completing Partial fractions, here's a quick recap of what we covered:

Skills covered

Mixed Practice

Exercises checked off

I'm Plex, here to help you understand this concept!
/
Algebra Skills
/
Partial fractions
3 by 3 Systems of equations
Partial fractions
Algebra Skills

Partial fractions

0 of 0 exercises completed

Decomposing rational expressions into partial fractions, including linear factors such as ​x+3A​+x−1B​​ and more complex forms, by cross-multiplying and solving for the unknown constants.

Linear partial fractions
AHL AA 1.11

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
​
Non-linear partial fractions
AHL AA 1.11

Aside from expressions of the form ​(x−c)(x−d)ax+b​, there are all sorts of types of partial fractions. For more complicated cases, exam questions will specify the form of the partial fractions decomposition, and you simply need to solve for the variables.


Solve for ​A​ and ​B​ in the equation ​x5−2x4+x−22x4+x​≡x4+1A​+x−2B​​


Cross multiplying:

​
x4+1A​+x−2B​=x5−2x4+x−2A(x−2)+B(x4+1)​
​

So

​
A(x−2)+B(x4+1)=2x4+x
​

We notice immediately that ​B=2, so

​
Ax−2A+2x4+2=2x4+x⇒A=1
​

Nice work completing Partial fractions, here's a quick recap of what we covered:

Skills covered

Mixed Practice

Exercises checked off

I'm Plex, here to help you understand this concept!

Generating starter questions...

1 free

Generating starter questions...

1 free