Module :MA0001NPFoundation Mathematics

Lecture Week 6

Rational ExpressionsP, Q, R, and S are polynomials

Addition

Operation

Multiplication

Subtraction

Division

P Q P Q

R R R

P Q P Q

R R R

P Q PQ

R S RS

P Q P S PS

R S R Q RQ

Notice the common denominator

Reciprocal and Multiply

Addition

bd

bcad

d

c

b

a

xx

2

31Simplify

)2(

)(3)2(1

xx

xx

)2(

32

xx

xx

)2(

22

xx

x

Addition

bd

bcad

d

c

b

a

y

x

x 2

3Simplify

)2(

))(()2(3

yx

xxy

xy

xy

2

6 2

Subtraction

bd

bcad

d

c

b

a

p

t

p 2

2

Simplify

)2)((

))(()2)(2(

pp

tpp

22

4

p

ptp

22

)4(

p

tp

p

t

2

)4(

Subtraction

bd

bcad

d

c

b

a

4

33

2

12

xxSimplify

)4)(2(

)33(2)12(4

xx

8

6648

xx

8

22 x

8

)1(2

x4

4

1x

Multiplication

bd

ac

d

c

b

a

5

5

3

x

xx

Simplify

)5(3

)5)((

x

xx

)5(3

52

x

xx

Multiplication

bd

ac

d

c

b

a

62

5

15

3

x

x

Simplify

)62)(15(

)5)(3(

x

x

)3)(2)(15(

)5)(3(

x

xFactorising (2x-6)Factorising (2x-6)

3

)2)(3(

1

6

1

Division

bc

ad

d

c

b

a

Simplify

14

4

7

2 yx

y

x

4

14

7

2

)4)(7(

)14)(2(

y

x

2

2

y

x

743

2xx

Solve

Solve 35

2

2

1

xx

Partial fractions Programme F7: Partial fractions

2( 1) 4( 3)2 43 1 ( 3)( 1)

2 2 4 12 ( 3)( 1)

10 2 ( 3)( 1)

x xx x x x

x xx x

xx x

Consider the following combination of algebraic fractions:

The fractions on the left are called the partial fractions of the fraction on the right.

Partial fractionsFind the partial fractions of the following

Programme F7: Partial fractions

28 28

6 8x

x x

Partial fractions

It is assumed that a partial fraction break down is possible in the form:

The assumption is validated by finding the values of A and B.

8 28( 2)( 4) 2 4

A Bxx x x x

Partial fractions

Partial fractions

To find the values of A and B the two partial fractions are added to give:

8 28( 2)( 4) 2 4

( 4) ( 2) ( 2)( 4)

A Bxx x x x

A x B xx x

Partial fractions Programme F7: Partial fractions

Since:

And since the denominators are identical the numerators must be identical as well. That is:

( 4) ( 2)8 28( 2)( 4) ( 2)( 4)

A x B xxx x x x

8 28 ( 4) ( 2)x A x B x

Partial fractions Programme F7: Partial fractions

Consider the identity:

Therefore:

8 28 ( 4) ( 2)x A x B x

Let 4 then 32 28 (0) (2) so 2x A B B Let 2 then 16 28 ( 2) (0) so 6x A B A

6 28 28( 2)( 4) 2 4x

x x x x

Therefore

Therefore:

Denominators with quadratic factors

A similar procedure is applied if one of the factors in the denominator is a quadratic. For example:

This results in:

2

2 215 2

5( 5)(3 4 2) 3 4 2

Bx Cx x Axx x x x x

2

2 215 2 (3 4 2) ( )( 5)

(3 ) (4 5 ) 2 5

x x A x x Bx C x

A B x A B C x A C

Denominators with quadratic factors

Equating coefficients of powers of x yields:

Three equations in three unknowns with solution:

2[ ] 15 3

[ ] 1 4 5

[ ] 2 2 5

x A B

x A B C

CT A C

4, 3 and 2 so that:A B C 2

2 2

3 215 2 45( 5)(3 4 2) 3 4 2

xx xxx x x x x

Denominators with repeated factorsRepeated factors in the denominator of the original fraction of the form:

give partial fractions of the form:

2( )ax b

2( )A B

ax b ax b

Partial fractions

Denominators with repeated factors

Partial fractions

Similarly, repeated factors in the denominator of the original fraction of the form:

give partial fractions of the form:

3( )ax b

2 3( ) ( )A B C

ax b ax b ax b

Partial fractions Programme F7: Partial fractions

Find the partial fractions for the following expressions:1.7x+18/(x+2)(x+3)

2.2x-7/x²+5x+4

3.-9/2x²+15x+18

4.5x-11/x²-5x+4

5.3x+11/2x²+3x-2

Partial fractions Programme F7: Partial fractions

Find the partial fractions for the following expressions:

6.x+21/2(2x+3)(3x-2)

7. x-35/x²-25

8.x-4/x²-6x+9

9.5x+4/-x²-x+2

10.12x-5/9x²-6x+1