Transcript C3.1 Algebra and functions 1

A2-Level Maths:
Core 3
for Edexcel
C3.1 Algebra and
functions 1
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Simplifying algebraic fractions
Contents
Simplifying algebraic fractions
Adding and subtracting algebraic fractions
Multiplying and dividing algebraic fractions
Improper fractions and polynomial division
Examination-style question
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Rational expressions
Remember, a rational number is any number that can be
written in the form a , where a and b are integers and b ≠ 0.
b
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 f ( x ) , where f(x) and g(x) are
g( x )
polynomials and g(x) ≠ 0.
For example,
3
x2
3 x +1
x2  2
x3  2
x2 + 3 x  4
For which values of x are each of the above expressions
undefined?
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Rational expressions
An algebraic fraction is undefined when the denominator is 0.
So,
3
is undefined when x + 2 = 0. That is, when x = –2.
x+2
3 x +1
2 – 2 = 0. That is, when x = ±√2.
is
undefined
when
x
x2  2
x3  2
2 + 3x – 4 = 0.
is
undefined
when
x
x2 + 3 x  4
We can factorize this to give (x + 4)(x – 1) = 0.
x3  2
So 2
is undefined when x = –4 or x = 1.
x + 3x  4
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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
2
2
28
=
42 3 3
The highest common factor of 28 and 42 is ___.
14
This fraction can therefore be written in its simplest terms by
dividing both the numerator and the denominator by 14.
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Simplifying algebraic fractions by cancelling
Algebraic fractions can be cancelled in the same way.
For example,
3
6a
6aa
3
=
=
3
8a
8  a  a  a 4a
2
4
When the numerator or the denominator contains more than
one term, we have to factorize before cancelling.
For example,
3 pq
Simplify
15 p  9 p 2
q
3 pq
3 pq
=
=
2
15 p  9 p
3 p(5  3 p ) 5  3 p
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Simplifying algebraic fractions by cancelling
3b  6
Simplify 2
b  2b
3b  6 3(b  2) 3
=
=
2
b  2b b(b  2) b
x2 + x  2
Simplify
x2  1
x 2 + x  2 ( x  1)( x  2) x  2
=
=
2
x 1
( x  1)( x  1) x  1
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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.
i.e.
a – b = –(b – a)
and
b – a = –(a – b )
For example,
3  y 2 ( y 2  3)
= 2
= –1
2
y 3
y 3
When cancelling, look for situations where a factor of –1 can
be taken out of a pair of brackets.
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Using additive inverses
14  7 x
Simplify 2
x  5x + 6
7(2  x )
14  7 x
=
x 2  5 x + 6 ( x  2)( x  3)
7(2  x )
=
(2  x )( x  3)
7
=
( x  3)
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,
3x  1
x
Simplify
x2
This can be simplified by multiplying the numerator and the
denominator by x.
3 x2  1
3 x  1x
=
2
x3
x
1 2
a
Simplify
a  3a
Multiply the numerator and the denominator by 3a:
1  2a
3a  6
3a  6
= 2
=
2
a
a  3 3a  a
4a 2
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Simplifying complex fractions
Simplify
3  42
x x
1  34x
To simplify this algebraic fraction we multiply the numerator
and the denominator by the lowest common multiple of x, x2
3x2
and 3x. That is ___.
3  42
9 x  12
x x
= 2
4
1 3 x 3 x  4 x
3(3 x + 4)
=
x(3 x + 4)
=
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3
x
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Contents
Adding and subtracting algebraic fractions
Simplifying algebraic fractions
Adding and subtracting algebraic fractions
Multiplying and dividing algebraic fractions
Improper fractions and polynomial division
Examination-style question
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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.
5 3
What is + ?
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.
This is the lowest common multiple (LCM) of their
denominators.
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Adding and subtracting fractions
The LCM of 6 and 4 is ___.
12
So we write,
5 3 10 + 9
+ =
6 4
12
19
=
12
7
=1 12
We apply the same method to add or subtract algebraic
fractions.
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Adding and subtracting fractions
2 3
Write 2 + as a single fraction in its lowest terms.
x
x
The LCM of x2 and x is ___.
x2
2 3 2+3
3xx
+
=
x2 x
x2
y x
 as a single fraction in its lowest terms.
Write
3x y
The LCM of 3x and y is ___.
3xy
y x y 22  3
3xx22
 =
3x y
3 xy
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Adding and subtracting fractions
2x
1
+
Write
as a single fraction in its lowest terms.
x + 3 2x + 6
2x
1
+
Start by factorizing where possible:
x + 3 2( x + 3)
The LCM of x + 3 and 2(x + 3) is ______.
2(x + 3)
2x
1
4x
1
+
=
+
x + 3 2( x + 3) 2( x + 3) 2( x + 3)
4 x +1
=
2( x + 3)
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Adding and subtracting fractions
2x + 8
Write 3  2
as a single fraction in its lowest terms.
x +5
2 x + 8 3( x 2 + 5) 2 x + 8
3 2
=
 2
2
x +5
x +5
x +5
3 x 2 +15  2 x  8
=
x2 + 5
Notice that this
becomes – 8.
3 x2  2 x  7
=
x2 + 5
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Contents
Multiplying and dividing algebraic fractions
Simplifying algebraic fractions
Adding and subtracting algebraic fractions
Multiplying and dividing algebraic fractions
Improper fractions and polynomial division
Examination-style question
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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.
3 12
What is ×
?
8 21
When multiplying two fractions, start by cancelling any
common factors in the numerators and denominators:
1
3
2
7
3 12
×
8 21
Then multiply the numerators and multiply the denominators:
1 3
3
× =
2 7 14
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Multiplying and dividing fractions
To divide by a fraction we multiply by its reciprocal.
7 14
What is ÷
?
9 15
This is equivalent to
1
5
3
2
7 15 1 5
×
= ×
9 14 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,
3 x  12
2
×
Simplify
2x + 4 x  4
Start by factorizing where possible:
3 x  12
2
3( x  4)
2
×
=
×
2 x + 4 x  4 2( x + 2) x  4
=
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3
x+2
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Multiplying and dividing algebraic fractions
x2  4
3x
×
Simplify
x + 2 2x  4
x2  4
3x
( x  2)( x  2)
3x
×
=
×
x + 2 2x  4
x+2
2( x  2)
3x
=
2
5 15
÷
Simplify
2p p
1
5 15
5
p
÷
=
×
=
2p p
2 p 15 6
3
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Multiplying and dividing algebraic fractions
14
7
÷ 2
Simplify
a+2 a a 6
14 a 2  a  6
14
7
×
÷ 2
=
7
a+2 a a 6 a+2
2
14 ( a + 2)( a  3)
=
×
a+2
7
= 2(a – 3)
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Contents
Improper fractions and polynomial division
Simplifying algebraic fractions
Adding and subtracting algebraic fractions
Multiplying and dividing algebraic fractions
Improper fractions and polynomial division
Examination-style question
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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.
This is called a mixed number.
For example, the improper fraction 29
can be converted to a
6
mixed number as follows:
29 24 + 5 24 5
=
=
+ =
6
6
6 6
5
4
6
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,
x3
x4
x3
x2  5
and
4 x2
( x  4)( x  2)
are improper algebraic fractions.
Suppose we have an improper fraction
f ( x)
.
g( x )
Dividing f(x) by g(x) will give us a quotient q(x) and a
remainder r(x), which gives us the identity:
f ( x)
r( x )
 q( x ) +
g( x )
g( x)
where the degree of f(x) ≥ the degree of g(x).
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Writing improper fractions in proper form
r( x )
We can think of the form q( x ) +
as being the algebraic
g( x )
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:
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.
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.
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Writing improper fractions in proper form
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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,
B
x+3
Write
in the form A +
.
x

1
x 1
x + 3 x  1+ 4
=
x 1
x 1
x 1
4
=
+
x 1 x 1
4
=1+
x 1
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Rewriting the numerator
What is the quotient and the remainder when 3x2 + 2x is
divided by x2 + 1?
We can write this in fraction form as:
3 x2 + 2 x
3( x 2 +1) + 2 x  3
=
2
x +1
x 2 +1
3( x 2 +1) 2 x  3
=
+ 2
2
x +1
x +1
2x  3
=3 + 2
x +1
So when 3x2 + 2x is divided by x2 + 1 the quotient is 3 and the
remainder is 2x – 3.
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Rewriting the numerator
Find the remainder when x3 is divided by x2 – 3.
We can write this in fraction form as:
x3
x3  3 x + 3 x
=
2
x 3
x2  3
x( x 2  3) + 3 x
=
x2  3
x( x 2  3)
3x
= 2
+ 2
x 3
x 3
3x
= x+ 2
x 3
The remainder is 3x.
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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 ( x)
r( x )
 q( x ) +
g( x )
g( x)
where q(x) is the quotient and r(x) is the remainder when f(x) is
divided by g(x).
What is x3 – 4x2 + 5 divided by x2 – 3?
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 identity
x3  4 x 2 + 5
Cx + D

Ax
+
B
+
x2  3
x2  3
Multiply through by x2 – 3:
x3  4 x 2 + 5  ( Ax + B )( x 2  3) + Cx + D
 Ax3  3 Ax + Bx 2  3 B + Cx + D
Equate the coefficients:
x 3:
A=1
B = –4
x 2:
x:
constants:
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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:
x3  4 x 2 + 5
3x  7
 x4+ 2
2
x 3
x 3
Alternatively, use long division:
x–4
x2 – 3 x3 – 4x2 + 0x + 5
x3 + 0x2 – 3x
– 4x2 + 3x + 5
– 4x2 +0x + 12
3x – 7
The quotient is x – 4 and the remainder is 3x – 7 so, as before:
x3  4 x2 + 5
3x  7
 x 4+ 2
2
x 3
x 3
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Contents
Examination-style question
Simplifying algebraic fractions
Adding and subtracting algebraic fractions
Multiplying and dividing algebraic fractions
Improper fractions and polynomial division
Examination-style question
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Examination-style question
Given that f ( x )  x 
3
9
+ 2
{ x  , x  1, x  2},
x2 x  x2
x2 + x  3
.
show that f ( x ) 
x +1
3
9
3
9
x
+ 2
=x
+
x2 x  x2
x  2 ( x  2)( x +1)
x( x  2)( x +1)
3( x +1)
9
=

+
( x  2)( x +1) ( x  2)( x +1) ( x  2)( x +1)
x3  x 2  2 x  3 x  3 + 9
=
( x  2)( x +1)
x3  x 2  5 x + 6
=
( x  2)( x +1)
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Examination-style question
Now divide x3 – x2 – 5x + 6 by x – 2:
x–2
So,
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x2 + x – 3
x3 – x2 – 5x + 6
x3 – 2x2
x2 – 5x
x2 – 2x
–3x + 6
–3x + 6
0
x3  x 2  5 x + 6 ( x  2)( x 2 + x  3)
=
( x  2)( x +1)
( x  2)( x +1)
x2 + x  3
as required.
=
x +1
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