module :ma0001np foundation mathematics lecture week 6
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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
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