revision partial fractions

DHS H2 MATHEMATICS : a revision package

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At the end of this revision, you should be able to:

Express a given algebraic fraction in terms of partial fractions

Use the cover-up rule

Express an improper algebraic fraction as aproper fraction before expressing it in partialfractions

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A complex algebraic fraction can be expressed asthe sum of 2 or more simpler fractionspartial

, known frac

astions

9 4

(3 5)( 2)

x

x x

3

3 5x

2

- 2x

the complex algebraic fraction its partial fractions

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The number and the form of the partial fractions of an F( )

algebraic fraction depends factor son i the n G( : G( )

)x

xx

2 2

F( ) 2 5Consider

G( ) 1 2 3

x x

x x x x

a non-repeated linear factor of the form ax b a repeated linear factor, i.e. ( 2) is repeated twicex

2a non-repeated quadratic factor of the form ax bx c

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In G( ), there is a:x

For every in G( ),

there will be a parti

non-repea

al fracti

ted l

on o

inear fac

f the f

tor

orm

xA

ax b

ax b

10 4

( 1)(3 2 )2

x

x x

3 2 2 1x

A

x

B

(2 (8

10 4

(3 2 1) 1 1 ))x x

x

x

23 1 12 8 1

B

x

A

x

C

x

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2

2For a in G( ), there'll

be partial fractions

rep

of

eated linear fa

the form

t

c or ax b xA B

ax b ax b

2

5 1

(2 1)

x

x

22 1 (2 1)

A B

x x

2(2 1

5 1

(7 4))x

x

x

22 1 (2 1 (7 4))

A

x

B

x

C

x

2

2For every in G( ),

there'll be a partial fraction of

quadratic fac

the form

tor xAx B

ax b

b

x c

ax x c

2(2 3 () 1)

x

xx

22 3 1

Ax B

x

C

x

22

1

5xx

2 2 5

C

x

A B

x

x

x

D

Note that x2 is a repeated linear factor (i.e. x repeated twice)

Note that x2 is a repeated linear factor (i.e. x repeated twice)

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We can determine these numerator constants by:

After writing out the form of the partial fractions, weneed to find the constants in the numerators of thesepartial fractions. For example,

2 2(2 3)( 1) 2 3 1

x Ax B C

x x x x

We want to find what values are A, B and C.

We want to find what values are A, B and C.

coveusi r-ung the p r ,ule like terms in on bothcompa sidr es, oring x suitablesubstitu values ting of x NEXT

The is used to find the numerator constant of a partial fraction corresponding to a non-repeated linear factor in G( )

cov

. F

er-up ru

or exam ,

le

plex

COVER-UP RULE COMPARING SUBSTITUTION

2 2

non-repeatedlinear factor

2 2

(( 1 1) 1) 1

x Bx C

x x

A

x x

We can use the cover-up rule to find as it isthe numerator of the partial fraction corresponding to t non-repeated linear factohe r 1.

A

x NEXT

The root of 1 is 1 (since if 1 0 1)...x x x

COVER-UP RULE COMPARING SUBSTITUTION

2 2

2 2

( 1)( 1) 1 1

x Bx C

x x x

A

x

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We "cover up" 1 in the algebraic fraction...x

Sub the root 1 into algebraic fraction to find :x A

2

2 1 22

1 1A

2 2

2 2

( 1)( 1) 1 1

A Bx C

xx x

x

x

Now to find and , we either compare like terms in , or substitute in suitable values of . But first, we "combine" the partial fractions on the RHS:

B C xx

COVER-UP RULE COMPARING SUBSTITUTION

2 2

2 2 2

( 1)( 1) 1 1

x

x x x x

Bx C

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2

2 2( 1)( 1) ( 1

2 2 2( 1) ( )( 1

)( 1)

)x x Bx C x

x x x x

...and equate the numerators on both sides:22 2 2( 1) ( )( 1)x x Bx C x

If you are quick, skip this step and go to the next step (i.e. equate the numerators)...

If you are quick, skip this step and go to the next step (i.e. equate the numerators)...

COVER-UP RULE COMPARING SUBSTITUTION

22 2 2( 1) ( )( 1)x x Bx C x

2Comparing the terms,x2 2 2( 0 (LHS) 2 (RHS))2 0 x x BB x

Comparing the constant terms,( 2 (LHS) 2 (RHS2 )2 )CC

2B

0C

We can now compare like terms in on both sides:x

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COVER-UP RULE COMPARING SUBSTITUTION

22 2 2( 1) ( )( 1)x x Bx C x

Sub 0 :x 22(0) 2 2(0 1) ( (0) )(0 1)B C

2 2 C

We can also substitute suitable values of to find and .

xB C

0C

Sub 2 :x 6 10 2B

2B

It is actually faster here to compare like terms in x to find B and C. So always try to use the faster method.

It is actually faster here to compare like terms in x to find B and C. So always try to use the faster method.

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This makes B disappear, leaving only C.This makes B disappear, leaving only C.

COVER-UP RULE COMPARING SUBSTITUTION

And finally, write out the partial fractions with thefound values for the numerator constants:

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2 2

2 2 2 2

( 1)( 1) 1 1

x x

x x x x

To check whether your partial fractions are correct, "combine" back the partial fractions on the RHS to see if you could obtain the algebraic fraction on the LHS.

To check whether your partial fractions are correct, "combine" back the partial fractions on the RHS to see if you could obtain the algebraic fraction on the LHS.

COVER-UP RULE rewind

COMPARING SUBSTITUTION

The can also be used to find the numerator of the partial fraction which denominator is the highest power of a repeated linear factor in G

c

( ). For example,

over-up rule

x

2

non-repeated repeated highest powerlinear factor linear factor of the repeated

linear

2 2

factor

11

2 1 12 1

x x A B

x xx x

C

x

We can use the cover-up rule to find (i.e. numerator of partial fraction corresponding to non-repeated linear factor)...

A

...and (i.e. numerator of partial fraction which denominator is the highest power of the repeated linear factor).

CNEXT

COVER-UP RULE rewind

COMPARING SUBSTITUTION

2

non-repeated repeated highest powerlinear factor linear factor of the repeated

linear

2 2

factor

11

2 1 12 1

x x A B

x xx x

C

x

The root of 1 is 1 (since if 1 0 1)...x x x

We "cover up" 1 in the algebraic fraction...x

Sub the root 1 into algebraic fraction to find C:x

211 1 13

1 2C

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The cover-up rule is actually a shortcut for substituting values of x. Can you see out why?

The cover-up rule is actually a shortcut for substituting values of x. Can you see out why?

COVER-UP RULE rewind

COMPARING SUBSTITUTION

2

2

11 3

( 2)( 1) 2 1 ( 1)

x x A B

x x x x x

Now find (using cover-up method) and (by comparing like terms or substituting suitable values):

A B

(1) 1, 2

(2) 1, 2

(3) 1, 2

(4) 1, 2

A B

A B

A B

A B

Click on correct

answer to proceed

COVER-UP RULE rewind

COMPARING SUBSTITUTION

2

2

11 3

( 2)( 1) 2 1 ( 1)

x x A B

x x x x x

Now find (using cover-up method) and (comparing like terms or substituting suitable values):

A B

(1) 1, 2

(2) 1, 2

(3) 1, 2

(4) 1, 2

A B

A B

A B

A B

Yes, this is the correct answer!

Yes, this is the correct answer!

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...can be expressed in partial fractions straightaway.

not fully factA) If its denominator is , we needto factorise it f

oi

risedrst:

22( 1)

4

( 3) 3 1

x A x

x

B C

x x x

2

4 4

( 3)( 1) ( 3)( 1)( 1) 3 1 1

x x A B C

x x x x x x x x

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not fully factorised

fully factorised

...can be expressed in partial fractions straightaway.

impropB) If the algebraic fraction is , we needto make it proper fi

errst:

is improper if degree of deF(

gree of G(x))

F(x) .G( )

x

x

For example, the following are improper:3 1

1

x

x

degree 3

degree 1 2 1

x

x

degree 1

degree 1 II

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F( )We can make an improper fraction proper

G( )using either:

x

x

Long division - if degree F( ) G( )x x

Splitting the numerator - if degree F( ) G( )x x

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LONG DIVISION SPLITTING NUMERATOR

3 23 4 5

1 2

x x x

x x

3 2

2

3 4 5

2

x x x

x x

2 3 2

3 2

2

2

2 3 4 5

(3 3 6 )

2 2 5

(2 2 4

1

2

)

3

4

x x x x x

x x x

x x

x x

x

x

( 1)( 23

42

)

1x

x xx

5 73 2 (cover-up rule)

3( 1) 3( 2)x

x x

properfraction now

properfraction now

improper fraction(deg of numerator

> deg of

denominator)

improper fraction(deg of numerator

> deg of

denominator)

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LONG DIVISION SPLITTING NUMERATOR

23 1

3 1

x

x x

2

2

3 1

2 3

x

x x

2

2

2

2 3

3 6 03 1xx x

x x

10 6

33 1

x

x x

"introduce" denominator in the

numerator

"introduce" denominator in the

numerator

improper fraction(deg of numerator = deg of denominator)

improper fraction(deg of numerator = deg of denominator)

split the numerator to create proper fractionsplit the numerator to create proper fraction

7 13 (cover-up rule)

3 1x x

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Which of the above fractions cannot be expressed in partial fractions straightaway:

2 2

2 2

2

Consider the following fractions:

2 5 2 5(A) (B)

( 4 5)( 1) (x 4 3)( 1)

1 9(C) (D)

( 1)( 3) ( 1)

x x

x x x x x

x x

x x x x

A, B B, C B, D C, DClick on correct

answer to proceed

Which of the above fractions cannot be expressed in partial fractions straightaway:

2 2

2 2

2

Consider the following fractions:

2 5 2 5(A) (B)

( 4 5)( 1) (x 4 3)( 1)

1 9(C) (D)

( 1)( 3) ( 1)

x x

x x x x x

x x

x x x x

A, B B, C B, D C, D

Bingo! Bingo!

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F( )To express in partial fractions, we need to:

G( )

x

x

determine the form of the partial fractions by looking at the factors in G( )x

determine the constants in the numerators of the partial fractions by:(A) using cover-up rule,(B) comparing like terms in , or(C) substituting suitable values of

xx

F( )first make sure is proper and G( ) is fully factorised

G( )

xx

x

Complete this assignment and submit to your tutor(your tutor will set a dateline)

Express the following in partial fractions:

2

2

5(A)

( 2)(3 1)

4 7(B)

(2 )(1 )

3 23 45(C)

( 3)

x

x x

x

x x

x x

x x