C4.1 Algebra and functions

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A2-Level Maths:
Core 4
for Edexcel
C4.1 Algebra and
functions
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Contents
Partial fractions
Partial fractions
Denominators with distinct linear factors
Denominators with a repeated linear factor
Improper fractions
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Partial fractions
We know that two or more algebraic fractions can be added or
subtracted to give a single fraction. For example:
2( x + 3)  ( x  5)
2
1

=
x 5 x+3
( x  5)( x + 3)
2x + 6  x + 5
=
( x  5)( x + 3)
x +11
=
( x  5)( x + 3)
Now, suppose we want to reverse the process.
x +11
In other words, suppose we are given
and
( x  5)( x + 3)
asked to express it as a sum of two separate fractions.
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Two distinct linear factors
x +11
This process is called expressing
in partial
( x  5)( x + 3)
fractions.
For example:
7x  1
Express
in partial fractions.
( x +1)( x  3)
The first step is to set up an identity.
The denominator of this fraction has two distinct linear
factors so let
7x  1
A
B

+
( x +1)( x  3) x +1 x  3
where A and B are constants to be found.
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Two distinct linear factors
7x  1
A
B

+
( x +1)( x  3) x +1 x  3
7x  1
A( x  3) + B( x +1)

( x +1)( x  3)
( x +1)( x  3)
7 x  1  A( x  3) + B( x +1)
1
There are now two ways to continue:
using suitable substitutions,
by equating coefficients.
To solve by substitution, we choose values of x that make the
brackets zero.
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Two distinct linear factors
Here substitute x = –1 into 1 :
7  1= A(4) + B(0)
4 A = 8
A=2
Now substitute x = 3 into 1 :
21 1= A(0) + B(4)
4 B = 20
B=5
Therefore
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7x  1
2
5

+
( x +1)( x  3) x +1 x  3
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Contents
Denominators with distinct linear factors
Partial fractions
Denominators with distinct linear factors
Denominators with a repeated linear factor
Improper fractions
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Two distinct linear factors
2x
Express
in partial fractions.
(3 x  2)(2 x  1)
2x
A
B

+
(3 x  2)(2 x  1) 3 x  2 2 x  1
Let
This can be simplified by multiplying through by (3x – 2)(2x –1):
2x  A(2x  1) + B(3 x  2)
This time if we use the substitution method we’ll have to
substitute fractional values for x.
Let’s multiply out the brackets and equate coefficients instead.
2 x  2 Ax  A + 3 Bx  2 B
2x  (2 A + 3B)x  A  2B
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Two distinct linear factors
Equate the coefficients of x:
2 = 2 A + 3B
1
0 =  A  2B
2
Now equate the constants:
1 + (2 × 2 ) gives:
2 = B
B = 2
Substitute this into 1 to find A:
2 = 2A  6
2A = 8
A=4
Therefore
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2x
4
2


(3 x  2)(2 x  1) 3 x  2 2 x  1
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Three distinct linear factors
9 x +1
Express
as a sum of partial fractions.
( x  3)( x +1)(2 x +1)
This time we have three distinct linear factors, so:
9 x +1
A
B
C

+
+
( x  3)( x +1)(2 x +1) x  3 x +1 2 x +1
Let
Multiply through by (x – 3)(x +1)(2x +1):
9 x +1  A( x +1)(2x +1) + B( x  3)(2 x +1) + C( x  3)( x +1) 1
To find A, substitute x = 3 into 1 :
27 +1= A(4)(7)
28 = 28 A
A =1
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Three distinct linear factors
We can find B by substituting x = –1 into 1 :
9 +1= B(4)(1)
8 = 4B
B = 2
We can now find C either by substituting x = 21 or by reverting to
the method of equating coefficients.
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Three distinct linear factors
To avoid awkward arithmetic involving fractions, let’s form an
equation in C by equating the constant terms in 1 .
9 x +1  A( x +1)(2x +1) + B( x  3)(2 x +1) + C( x  3)( x +1) 1
1= A  3 B  3C
1= 1+ 6  3C
We could also equate
the coefficients of
x2 or x.
3C = 6
C =2
9 x +1
1
2
2


+
Therefore
( x  3)( x +1)(2 x +1) x  3 x +1 2 x +1
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Contents
Denominators with a repeated linear factor
Partial fractions
Denominators with distinct linear factors
Denominators with a repeated linear factor
Improper fractions
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Denominators with a repeated linear factor
6 x2  x  2
Suppose we wish to express
in partial fractions.
2
( x + 4)( x  3)
This is an example of a fraction whose denominator contains a
repeated linear factor.
In this case, the partial fractions will be of the form:
6 x2  x  2
A
B
C

+
+
2
( x + 4)( x  3)
x + 4 x  3 ( x  3)2
We can now find A, B and C using a combination of
substitution and equating the coefficients.
6 x2  x  2
A( x  3)2  B( x  4)( x  3)  C ( x  4)

2
( x + 4)( x  3)
( x + 4)( x  3)2
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Denominators with a repeated linear factor
6 x 2  x  2  A( x  3)2 + B( x + 4)( x  3) + C( x + 4)
1
Substitute x = –4 into 1 :
6( 4)2  ( 4)  2 = A( 7)2
98 = 49 A
A=2
Substitute x = 3 into 1 :
6(9)  3  2 = C(7)
49 = 7C
C =7
To find B we can switch to the method of comparing
coefficients.
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Denominators with a repeated linear factor
Equate the coefficients of x2 in 1 :
6 = A+ B
But A = 2 so:
B=4
Therefore
6 x2  x  2
2
4
7

+
+
2
( x + 4)( x  3)
x + 4 x  3 ( x  3)2
13 x +12
Express 2
as a sum of partial fractions.
x (4  x )
13 x +12 A B
C

+
+
x2 (4  x) x x2 4  x
Let
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Denominators with a repeated linear factor
Multiply through by x2(4 – x):
13 x +12  Ax(4  x ) + B(4  x ) + Cx 2
1
Substitute x = 0 into 1 :
12 = 4B
B=3
Substitute x = 4 into 1 :
64 = 16C
C=4
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Denominators with a repeated linear factor
We can find A by comparing the coefficients of x2.
0 = A+ C
A=C
But C = 4 so:
A=4
Therefore
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13 x +12 4 3
4
 + 2+
2
x (4  x) x x
4x
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Contents
Improper fractions
Partial fractions
Denominators with distinct linear factors
Denominators with a repeated linear factor
Improper fractions
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Improper fractions
Remember, an algebraic fraction is called an improper
fraction if the degree of the polynomial is equal to, or
greater than, the degree of the denominator.
To express an improper fraction in partial fractions we start by
expressing it in the algebraic equivalent of mixed number form.
Any proper fractions contained in this form can then be
expressed in partial fractions.
2 x 2  3 x +13
Express 2
in partial fractions.
x  2 x  15
We can either use long division to divide 2x2 – 3x + 13 by
x2 – 2x – 15 or we can set up an identity as follows:
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Improper fractions
Start by factorizing the denominator.
2 x2  3 x +13 2 x 2  3 x +13

2
x  2 x  15 ( x + 3)( x  5)
The numerator and the denominator are both of degree 2 and
so they will divide to give a constant, A.
The part that is a proper fraction will have two distinct linear
factors. So we can let
2 x 2  3 x +13
B
C
 A+
+
( x + 3)( x  5)
( x + 3) ( x  5)
Multiply through by (x + 3)(x – 5):
2 x 2  3 x +13  A( x + 3)( x  5) + B( x  5) + C( x + 3)
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Improper fractions
Substitute x = –3 into 1 :
18 + 9 +13 = 8B
8 B = 40
B = 5
Substitute x = 5 into 1 :
50  15 +13 = 8C
8C = 48
C =6
We can find A by equating the coefficients of x2 in 1 .
A=2
Therefore
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2 x 2  3 x +13
5
6
 2
+
2
x  2 x  15
( x + 3) ( x  5)
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