1 Algebraic fractions
In this chapter you will learn how to simplify rational expressions.
1.1
You can treat algebraic fractions in exactly the same way as numerical ones. You
can cancel them down by finding factors that are common to both the numerator
and the denominator.
Example 1
Find the simplest forms of the fractions
x3
x2
a 1260
b c 2x 6
3x 8
a
b
c
16
44
20 5 4
54
x3
1 (x 3)
2x 6 2 (x 3)
21
x2
3x 8
Find a factor that is common to both the
numerator and the denominator and cancel
down.
Write both the numerator and the
denominator as products before cancelling
by (x 3).
This is the simplest form as there are no
common factors. Remember you cannot
cancel over addition, e.g.
1x
21 1 1
27 ✗
3x 84 3 4
When your algebraic expression has fractions in the numerator or denominator, it is sensible
to multiply both by the same ‘number’ to create an equivalent fraction.
Example 2
1
2x + 1
Simplify 1
2
3x + 3
1
x 1
2
1
2
x 3
3
1
(2x 1) 6
(31x 32) 6
3x 6
2x 4
3(x 2)
2(x 2)
32
The LCM of 2 and 3 is 6, so multiply
numerator and denominator by 6.
Factorise numerator and denominator.
Cancel any common factors.
1
CHAPTER
1
Example 3
Simplify the expressions:
1
x x
b x1
x2 1
a x2 4x 3
a
x2 1
(x 1)(x 1)
2
x 4x 3 (x 1)(x 3)
(x 1)(x 1)
(x 1)(x 3)
x1
x3
Factorise both numerator and denominator.
Cancel terms that are equal.
x x
x
x 1
1
2
b
Multiply numerator and denominator by x
to remove the fraction.
(x 1) x
x(x 1)
(x 1)(x 1)
x(x 1)
x1
x
x 1
OR x x
1
1 x
Factorise and cancel common factors of
(x 1).
By dividing throughout by x.
Exercise 1A
1
Simplify:
4x 4
a x1
2x 1
b 6x 3
x4
c x2
x 2
d 4x 2
4x 2y
e 6x 3y
a3
f a6
5p 5q
g 10p 10q
2a b
h 2a 4b
i x2 3x
x2 3x
j x2 9
x2 5x 4
k x2 8x 16
x3 2x2
l x2 4
x2 4
m
x2 4
x2
n 2
x 5x 6
2x2 5x 3
o 2x2 7x 4
3x2 x 2
q 1
1
2 x 3
x2 5x 6
r 1
3 x 2
1
1
2 x2 x 4
p 1 2
3
4 x 2 x 2
2
1
x2
Algebraic fractions
1.2
You multiply fractions together by finding the product of the numerator and
dividing by the product of the denominator.
Example 4
Calculate:
a
1
3
2
5
a
a c
b b d
3
1
2 5
13
25
130
b
a c
b d
ac
bd
ac
bd
Multiply numerators.
Multiply denominators.
When there are factors common to both the numerator and the denominator cancel down
first.
Example 5
Simplify the following products:
a c
x1
3
a 35 59
b c 2
x 1
b a
2
1
1
3 8
11
a 93 1 3
15
31
b
c
c 1c
b a1 b 1
c
b
x1
3
2
2
x 1
3
x1
(x 1)(x 1)
2
1
x1
3
2
(x 1)(x 1)
1
3
2(x 1)
1a
Cancel any common factors and multiply
numerators and denominators.
Cancel any common factors and multiply
numerators and denominators.
Factorise (x2 1).
Cancel any common factors and multiply
numerators and denominators.
To divide by a fraction multiply by the reciprocal of that fraction.
3
CHAPTER
1
Example 6
Divide 56 by 13.
5
1
5 31
1
6 3 26
51
21
Turn divisor upside down and multiply by it.
Cancel the common factor 3.
Multiply numerators and denominators.
52
Example 7
Simplify:
a a
a b c
a
b
x 2 3x 6
b x 4 x2 16
a a 1a c
b c b a1
1c
b1
c
b
Turn divisor upside down and multiply by it.
Cancel the common factor a.
Multiply numerators and denominators.
x 2 3x 6
x 4 x2 16
x 2 x2 16
x 4 3x 6
x 2 (x 4)(x 4)
3(x 2)
x4
1
1
x 2 (x 4)(x 4)
1 3(x 2)1
x4
x4
3
Turn divisor upside down and multiply by it.
Factorise x2 16 (difference of two squares).
Cancel any common factors.
Exercise 1B
1
Simplify:
a a
a d c
3 6
d x
4
x
a2 c
b a
c
x
4
e xy
y
2 x
c x 4
2r2 4
f 3
r
5
Algebraic fractions
1
g (x 2) 2
x 4
1
a2 9
h 2
a 6a 9
2
x2 3x y 1
i y2 y
x
j y 3 y2 4y 3
x2 2x3 6x2
k x2 3x
3
4x2 25
2x 5
l 8
4x 10
x2 5x
x3
m x2 10x 25 x2 3x
3y 6
3y2 4y 4
n 10
15
y
y2
4
x2 2xy y2
o 2
2
(x y)
1.3
You need to have the same denominator when you add or subtract numeric or
algebraic fractions.
Example 8
Add 13 to 34.
1
3
3
4
4
4
3
3
4
9
12
12
13
12
The lowest common multiple of 3 and 4 is 12.
Remember the lowest
common multiple of 3
and 4 is the smallest
number into which 3 and
4 both divide.
Example 9
Add 13 to 152.
4
1
5
12
3
4
4
5
12
12
9
12
3
4
The lowest common multiple of 3 and 12 is
12.
Cancel down.
5
CHAPTER
1
Example 10
Simplify the following fractions:
3
4x
a
a b
b 2
x1 x 1
x
a
a
b ax b1
x
b
Write b as .
1
x
x
ax bx
x
The LCM is x.
a x bx
b
3
x24x
x1
1
3
4x
(x (x 1)(x
1)
1)
(x 1)
(x 1)(x 1)
(x 1)
(x 1)
(x
1
3(x 1)
4x
(x 1)(x 1) (x 1)(x
1)
3(x 1) 4x
Factorise x2 1 to (x 1)(x 1).
The LCM of (x 1) and (x 1)(x 1) is
(x 1)(x 1).
Simplify the numerator: 3x 3 4x x 3.
(x 1)(x 1)
x 3
(x 1)(x 1)
Exercise 1C
1
Simplify:
1 1
a p q
a
b 1
b
1
1
c 2x x
3 1
d 2 3
1
e 4x 8x
f 1
1
g x2 x1
2
1
h x3 x2
i
3x
1
j 2 (x 4)
(x 4)
1
1
k 2(x 3) 3(x 1)
2
1
l 2
x 2x 1 x 1
3
2
m
2
2
x 3x 2 x 4x 4
2
3
n 2
2
a 6a 9 a 4a 3
2
3
o 2 2
y x
yx
x2
x1
p 2
2
x x 12 x 5x 6
3x 1
2
4
q 3 2 (x 2)
(x 2)
(x 2)
x
6
x
x
y
y
x
1
(
3
x 2) 12(x 3)
Algebraic fractions
1.4
You can divide two algebraic fractions. As with ordinary numbers, an improper
fraction has a larger numerator than denominator, e.g.
12
5
Improper fraction
2
2
5
Mixed number
An improper algebraic fraction is one whose numerator has a degree equal to or
larger than the denominator, e.g.
x2 5x 8
x3 8x 5
and x2
x2 2x 1
are both improper fractions because the numerator has a larger degree than the
denominator. You can change them into mixed number fractions either by long
division or by using the remainder theorem.
Example 11
Divide x3 x2 7 by x 3 by using long division.
Include all coefficients, including ‘0’.
x2 4x 12
x 3 x3 x2 0x 7
x3 3x2
4x2 0x
4x2 12x
12x 7
12x 36
29
3
2
x x 7
Therefore x3
x2 4x 12 remainder 29
29
x2 4x 12 x3
Divide x into x3: it divides in x2 times.
Multiply the divisor by x2 then subtract.
Divide x into 4x2: it divides in 4x times.
Multiply the divisor by 4x then subtract.
Divide x into 12x: it divides in 12 times.
Multiply the divisor by 12 then subtract.
How many whole times (x 3) divides into
x3 x2 7.
29
The remainder is because you are
x3
dividing by (x 3).
The remainder theorem:
Any polynomial F(x) can be put in the form
F(x) Q(x) divisor remainder
where Q(x) is the quotient and is how many times the divisor
divides into the function.
This is a more general
version of the remainder
theorem you met in Book
C2. It allows you to divide
by a quadratic expression.
7
CHAPTER
1
Example 12
Divide x3 x2 7 by x 3 by using the remainder theorem.
Set up the identity
F(x) Q(x) divisor remainder
x3 x2 7 (Ax2 Bx C)(x 3) D
and solve to find the constants A, B, C and D.
Let x 3
Let x 0
27 9 7 (9A 3B C) 0 D
D 29
0 0 7 (A 0 B 0 C) 3 D
7 3C D
7 3C 29
3C 36
C 12
Compare coefficients in x3 1 A
Compare coefficients in x2
1 3 A B
1 3 B
B 4
Therefore
x3 x2 7
(1x2 4x 12)(x 3) 29
x3 x2 7
29
⇒ x2 4x 12 x3
x3
As the divisor is a linear
expression and F(x) is a
cubic polynomial then
Q(x) must be a quadratic
and the remainder must
be a constant.
This is true for all x, so
you can substitute into
the RHS and LHS to work
out the values of A, B, C
and D.
Put x 3 into both
sides to give an
equation in D only.
Substitute D 29 and
x 0 to give an
equation in C only.
Because this is an equivalence
relation you can compare
coefficients of terms in x3 and
x2 on the RHS and LHS of the
equation
In x3 : LHS
RHS
In x2 : LHS
RHS
x3
Ax3
x2
(3A B)x2
Substitute A 1.
(x 3)
Example 13
Divide x4 x3 x 10 by x2 2x 3.
Method 1. Using long division.
x2 x 5
2
x 2x 3 x4 x3 0x2 x 10
x4 2x3 3x2
x3 3x2 x
x3 2x2 3x
5x2 2x 10
5x2 10x 15
12x 5
8
All coefficients need to be included.
x2 goes into x4, x2 times. Multiply x2 by
(x2 2x 3) and subtract.
x2 goes into x3, x times. Multiply x by
(x2 2x 3) and subtract.
x2 goes into 5x2, 5 times. Multiply 5 by
(x2 2x 3) and subtract.
Since the degree of (12x + 5) is smaller than
(x2 2x 3) this is the remainder.
Algebraic fractions
Therefore
x4 x3 x 10
x2 2x 3
x2 x 5 remainder 12x 5
12x 5
x2 x 5 x2 2x 3
Write the remainder as a ‘fraction’.
Method 2. Using the remainder theorem.
F(x) Q(x) divisor remainder
x4 x3 x 10
(Ax2 Bx C)(x2 2x 3) Dx E
Now solve for A, B, C, D and E.
As the divisor is a quadratic expression and
F(x) has a power of 4 then Q(x) must be a
quadratic and the remainder must be a
linear expression.
As x2 2x 3 (x 3)(x 1) we should
start by substituting x 1, x 3 then
x 0.
Exercise 1D
1
2
Express the following improper fractions in ‘mixed’ number form by:
i using long division
ii using the remainder theorem
x3 2x2 3x 4
a x1
2x3 3x2 4x 5
b x3
x3 8
c x2
2x2 4x 5
d x2 1
8x3 2x2 5
e 2x2 2
4x3 5x2 3x 14
f x2 2x 1
x4 3x2 4
g x2 1
x4 1
h x1
2x4 3x3 2x2 4x 6
i x2 x 2
Find the value of the constants A, B, C, D and E in the following identity:
3x4 4x3 8x2 16x 2 (Ax2 Bx C)(x2 3) Dx E
Mixed exercise 1E
1
Simplify the following fractions:
x2 2x 1
b 4x 4
x2 x x 1
c 2
4
a4
e a8
b2 4b 5
f b2 2b 3
a 4
3
3
4
b 2y
y
x1 x2
c 2
3
x2 5x 6
d x1
x3 7x 1
e x2
x4 3
f x2 1
ab
c2
a 2
c
a
1
x 2
x
d x1
2
Simplify:
x
x
9
CHAPTER
1
3
Find the value of the constants A, B, C and D in the following identity:
x3 6x2 11x 2 (x 2)(Ax2 Bx C) D
4
3
12
f(x) x 2
x 1 x 2x 3
{x , x 1}
x2 3x 3
Show that f(x) x3
5
x4 2
C
Show that x2 B for constants B and C, which should be found.
2
2
x 1
x 1
6
4x3 6x2 8x 5
D
Show that can be put in the form Ax2 Bx C . Find the
2x 1
2x 1
E
values of the constants A, B, C and D.
Summary of key points
1
Algebraic fractions can be simplified by cancelling down. To do this the numerators and
denominators must be fully factorised first.
2
If the numerator and denominator contain fractions then you can multiply both by the
same number (the lowest common multiple) to create an equivalent fraction.
3
To multiply fractions, you simply multiply the numerators and multiply the
denominators. If possible cancel down first.
4
To divide two fractions, multiply the first fraction by the reciprocal of the second fraction.
5
To add (or subtract) fractions each fraction must have the same denominator. This is done
by finding the lowest common multiple of the denominators.
6
When the numerator has the same or higher degree than the denominator, you can divide
the terms to produce a ‘mixed’ number fraction. This can be done either by using long
division or by using the remainder theorem:
F(x) Q(x) divisor remainder
where Q(x) is the quotient and is how many times the divisor divides into the function.
10