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Simplifying algebraic fractions
Adding and subtracting algebraic fractions
Multiplying and dividing algebraic fractions
Improper fractions and polynomial division
Examination-style question
Co
nte
nts
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Simplifying algebraic fractions
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Rational expressions
Remember, a rational number is any number that can be written in the form , where a and b are integers and b ≠ 0.
2
3
x 2
3 +1
2
x
x
3
2
2
+ 3 4
x
x x
ab
Numbers written in this form are often called fractions.
In algebra, a rational expression is an algebraic fraction that can be written in the form , where f(x) and g(x) are polynomials and g(x) ≠ 0.
fg( )( )xx
For example,
For which values of x are each of the above expressions undefined?
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Rational expressions
That is, when x = –2. is undefined when x + 2 = 0.
3
+ 2x
That is, when x = ±√2. is undefined when x2 – 2 = 0.2
3 +1
2
x
x
We can factorize this to give (x + 4)(x – 1) = 0.
is undefined when x2 + 3x – 4 = 0.3
2
2
+ 3 4
x
x x
So is undefined when x = –4 or x = 1.3
2
2
+ 3 4
x
x x
An algebraic fraction is undefined when the denominator is 0. So,
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Simplifying fractions by cancelling
When the numerator and the denominator of a numerical fraction contain a common factor, the fraction can be simplified by cancelling.
For example, consider the fraction
This fraction can therefore be written in its simplest terms by dividing both the numerator and the denominator by 14.
The highest common factor of 28 and 42 is ___. 14
28
42
2
3
2=
3
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Simplifying algebraic fractions by cancelling
Algebraic fractions can be cancelled in the same way.
For example,2
3
6=
8
a
a
6= =
8
a a
a a a
3
4
3=
4a
When the numerator or the denominator contains more than one term, we have to factorize before cancelling. For example,
Simplifypq
p p 2
3
15 9
2
3=
15 9
pq
p p3
=3 (5 3 )
pq
p p 5 3
q
p
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Simplifying algebraic fractions by cancelling
2
3 6=
2
b
b b
3( 2)=
( 2)
b
b b
3
b
Simplifyx x
x
2
2
+ 2
1
Simplify 2
3 6
2
b
b b
2
2
+ 2=
1
x x
x
( 1)( 2)=
( 1)( 1)
x x
x x
2
1
x
x
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Using additive inverses
When manipulating algebraic fractions it is helpful to remember that an expression of the form a – b is the additive inverse of the expression b – a.
a – b = –(b – a)
and
b – a = –(a – b )
For example,
When cancelling, look for situations where a factor of –1 can be taken out of a pair of brackets.
2
2
3=
3
y
y
2
2
( 3)=
3
y
y
–1
i.e.
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Using additive inverses
Simplify 2
14 7
5 + 6
x
x x
2
14 7=
5 + 6
x
x x
7(2 )
( 2)( 3)
x
x x
7(2 )=
(2 )( 3)
x
x x
7=
3 x
7=
( 3)x
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Simplifying complex fractions
Sometimes the numerator or the denominator of an algebraic fraction contains another fraction. For example,
Simplify2
13 xx
x
This can be simplified by multiplying the numerator and the denominator by x.
2
13=xx
x
2
3
3 1x
x
Simplify2
3
1 aaa
Multiply the numerator and the denominator by 3a:2
3
1=a
aa
2 2
3 6
3
a
a a
2
3 6=
4
a
a
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Simplifying complex fractions
Simplify
To simplify this algebraic fraction we multiply the numerator and the denominator by the lowest common multiple of x, x2 and 3x. That is ___.3x2
2
9 12
3 4
x
x x
3(3 + 4)=
(3 + 4)
x
x x
3=
x
23 4
43
=1x x
x
23 4
431
x x
x
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Simplifying algebraic fractions
Adding and subtracting algebraic fractions
Multiplying and dividing algebraic fractions
Improper fractions and polynomial division
Examination-style question
Co
nte
nts
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Adding and subtracting algebraic fractions
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Adding and subtracting fractions
Before looking at the addition and subtraction of algebraic fractions, let’s recall the method used for numerical fractions.
This is the lowest common multiple (LCM) of their denominators.
What is ?5 3
+6 4
Before we can add these two fractions we have to write them as equivalent fractions over a common denominator.
It is best to use the lowest common denominator of the two fractions.
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Adding and subtracting fractions
The LCM of 6 and 4 is ___.
So we write,5 3 10 + 9
+ =6 4 12
10 9
19=
12
12
We apply the same method to add or subtract algebraic fractions.
712=1
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2 2
2 3 2 + 3+ =
x
x x x
Adding and subtracting fractions
Write as a single fraction in its lowest terms.2
2 3+
x x
The LCM of x2 and x is ___.x2
2 3x
Write as a single fraction in its lowest terms.3
y x
x y
2 23=
3 3
y x y x
x y xy
The LCM of 3x and y is ___.3xy
y2 3x2
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Adding and subtracting fractions
Write as a single fraction in its lowest terms.2 1
++ 3 2 + 6
x
x x
The LCM of x + 3 and 2(x + 3) is ______.2(x + 3)
Start by factorizing where possible:2 1
++ 3 2( + 3)
x
x x
2 1+ =
+ 3 2( + 3)
x
x x
4 1+
2( + 3) 2( + 3)
x
x x
4 +1=
2( + 3)
x
x
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Adding and subtracting fractions
Write as a single fraction in its lowest terms.2
2 + 83
+ 5
x
x
2
2 + 83 =
+ 5
x
x
2
2 2
3( + 5) 2 + 8
+ 5 + 5
x x
x x
2
2
3 +15 2 8=
+ 5
x x
x
Notice that this becomes – 8.
2
2
3 2 7=
+ 5
x x
x
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Simplifying algebraic fractions
Adding and subtracting algebraic fractions
Multiplying and dividing algebraic fractions
Improper fractions and polynomial division
Examination-style question
Co
nte
nts
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Multiplying and dividing algebraic fractions
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Multiplying and dividing fractions
Before looking at the multiplication and division of algebraic fractions, let’s recall the methods used for numerical fractions.
What is ?3 12
×8 21
When multiplying two fractions, start by cancelling any common factors in the numerators and denominators:
3 12×
8 21
1
7
3
2
Then multiply the numerators and multiply the denominators:
1 3× =
2 7
3
14
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Multiplying and dividing fractions
To divide by a fraction we multiply by its reciprocal.
What is ?7 14
÷9 15
This is equivalent to7 15
× =9 14
1
2
5
3
1 5×
3 2
5=
6
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Multiplying and dividing algebraic fractions
We can apply the same methods to the multiplication and division of algebraic fractions. For example,
Simplify3 12 2
×2 + 4 4
x
x x
Start by factorizing where possible:
3 12 2× =
2 + 4 4
x
x x
3( 4) 2×
2( + 2) 4
x
x x
3=
+ 2x
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Multiplying and dividing algebraic fractions
Simplify2 4 3
×+ 2 2 4
x x
x x
2 4 3× =
+ 2 2 4
x x
x x
( 2)( 2) 3×
+ 2 2( 2)
x x x
x x
3=
2
x
Simplify5 15
÷2 p p
5 15÷ =
2 p p
5× =
2 15
p
p
1
63
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Multiplying and dividing algebraic fractions
Simplify 2
14 7÷
+ 2 6a a a
2
14 7÷ =
+ 2 6a a a
214 6×
+ 2 7
a a
a
14 ( + 2)( 3)= ×
+ 2 7
a a
a
2
= 2(a – 3)
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Simplifying algebraic fractions
Adding and subtracting algebraic fractions
Multiplying and dividing algebraic fractions
Improper fractions and polynomial division
Examination-style question
Co
nte
nts
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Improper fractions and polynomial division
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Improper fractions and mixed numbers
Remember, a numerical fraction is called an improper fraction if the numerator is larger than the denominator.
Improper fractions are usually simplified by writing them as a whole number plus a proper fraction.
For example, the improper fraction can be converted to a mixed number as follows:
296
29 24 + 5= =
6 6
24 5+ =
6 6
5
64
This is called a mixed number.
When 29 is divided by 6, 4 is the quotient and 5 is the remainder.
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Improper algebraic fractions
An algebraic fraction is called an improper fraction when the numerator is a polynomial of degree greater than, or equal to, the degree of the denominator.
For example,
are improper algebraic fractions.
3
2 5
x
x
24
( 4)( 2)
x
x x 3
4
x
x
and
Suppose we have an improper fraction . fg( )( )xx
Dividing f(x) by g(x) will give us a quotient q(x) and a remainder r(x), which gives us the identity:
f rq
g g
( ) ( )( )+
( ) ( )
x xx
x x
where the degree of f(x) ≥ the degree of g(x).
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Writing improper fractions in proper form
We can think of the form as being the algebraic
equivalent of mixed number form. It is a polynomial plus a proper fraction.If the degree of f(x) is n and the degree of g(x) is m then:
An improper algebraic fraction can be written in proper form by:
rewriting the numerator.
writing an appropriate identity to equate the coefficients.
using long division to divide the numerator by the denominator.
rq
g
( )( )+
( )
xx
x
The degree of the quotient q(x) will be equal to n – m. The degree of the quotient q(x) will be equal to n – m.
The degree of the remainder r(x) will be less than m.The degree of the remainder r(x) will be less than m.
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Writing improper fractions in proper form
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Write in the form A + .+ 3
1
x
x 1
B
x
Rewriting the numerator
A useful technique for writing improper fractions in proper form is to look for ways to rewrite the numerator so that part of it can be divided by the denominator. For example,
+ 3
1
x
x 1+ 4
=1
x
x
1 4= +
1 1
x
x x
4=1+
1x
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Rewriting the numerator
2
2 2
3( +1) 2 3= +
+1 +1
x x
x x
2
2 3= 3 +
+1
x
x
We can write this in fraction form as:2
2
3 + 2
+1
x x
x
2
2
3( +1)+ 2 3=
+1
x x
x
So when 3x2 + 2x is divided by x2 + 1 the quotient is 3 and the remainder is 2x – 3.
What is the quotient and the remainder when 3x2 + 2x is divided by x2 + 1?
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Rewriting the numerator
2
2
( 3)+ 3=
3
x x x
x
2
2 2
( 3) 3= +
3 3
x x x
x x
We can write this in fraction form as:
3
2 3
x
x
3
2
3 + 3=
3
x x x
x
The remainder is 3x.
2
3= +
3
xx
x
Find the remainder when x3 is divided by x2 – 3.
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Constructing an identity
When the numerator cannot easily be manipulated to give an expression of the required form, we can write an identity using:
f rq
g g
( ) ( )( )+
( ) ( )
x xx
x x
What is x3 – 4x2 + 5 divided by x2 – 3?
where q(x) is the quotient and r(x) is the remainder when f(x) is divided by g(x).
Let the quotient be Ax + B. (It must be linear because the degree of the dividend minus the degree of the divisor is 1).
Let the remainder be Cx + D. (Its degree must be less than 2.)
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Constructing an identity
This gives us the identity3 2
2 2
4 + 5 ++ +
3 3
x x Cx DAx B
x x
Multiply through by x2 – 3:
3 2 24 + 5 ( + )( 3)+ +x x Ax B x Cx D 3 23 + 3 + +Ax Ax Bx B Cx D
Equate the coefficients:
x3: A = 1
x2:
x:
constants:
B = –4
C – 3A = 0 C = 3
D – 3B = 5 D + 12 = 5
D = –7
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Constructing an identity
We can now substitute these values into the original identity to give:
3 2
2 2
4 + 5 3 74 +
3 3
x x xx
x x
Alternatively, use long division:
x3 – 4x2 + 0x + 5x2 – 3x3 + 0x2 – 3x
– 4x2 + 3x + 5
x
– 4x2 +0x + 123x – 7
– 4
The quotient is x – 4 and the remainder is 3x – 7 so, as before:3
2 2
2 3 74 +
3
4 + 5
3
x x
x
xx
x
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Co
nte
nts
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Simplifying algebraic fractions
Adding and subtracting algebraic fractions
Multiplying and dividing algebraic fractions
Improper fractions and polynomial division
Examination-style question
Examination-style question
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Examination-style question
Given that
show that
f2
3 9( ) + { , 1, 2},
2 2x x x x x
x x x
f2 + 3
( ) .+1
x xx
x
2
3 9 3 9+ = +
2 2 2 ( 2)( +1)x x
x x x x x x
( 2)( +1) 3( +1) 9= +
( 2)( +1) ( 2)( +1) ( 2)( +1)
x x x x
x x x x x x
3 2 2 3 3 + 9
=( 2)( +1)
x x x x
x x
3 2 5 + 6=
( 2)( +1)
x x x
x x
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Examination-style question
Now divide x3 – x2 – 5x + 6 by x – 2:
x3 – x2 – 5x + 6x – 2 x3 – 2x2
x2 – 5x
+ x
x2 – 2x–3x + 6
– 3
–3x + 60
x2
So,3 2 25 + 6 ( 2)( + 3)
=( 2)( +1) ( 2)( +1)
x x x x x x
x x x x
x x
x
2 + 3=
+1as required.