Download A2-Level Maths: Core 3 for Edexcel

A2-Level Maths:
Core 3
for Edexcel
C3.1 Algebra and
functions 1
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© Boardworks Ltd 2006
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
2 of 37
© Boardworks Ltd 2006
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.a
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
, where f(x) and g(x) are polynomials and g(x)f ≠( x0.)
g( x )
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?
Rational expressions
An algebraic fraction is undefined when the denominator is 0. So,
3 is undefined when x + 2 = 0.
x+2
3 x +1
is undefined when x2 – 2 = 0.
x2  2
x3  2is undefined when x2 + 3x – 4 = 0.
x2 + 3 x  4
We can factorize this to give (x + 4)(x – 1) = 0.
So
x3 is2undefined when x = –4 or x = 1.
x2 + 3 x  4
That is, when x = –2.
That is, when x = ±√2.
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.
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,
Simplify
3 pq
15 p  9 p 2
q
3 pq
3 pq
=
=
2
15 p  9 p
3 p(5  3 p ) 5  3 p
Simplifying algebraic fractions by
cancelling
Simplify
3b  6
b 2  2b
3b  6 3(b  2) 3
=
=
2
b  2b b(b  2) b
Simplify
x2 + x  2
x2  1
x 2 + x  2 ( x  1)( x  2) x  2
=
=
2
x 1
( x  1)( x  1) x  1
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) –1
= 2
=
2
y 3
y 3
When cancelling, look for situations where a factor of –1 can be taken out of a pair of
brackets.
Using additive inverses
Simplify
14  7 x
x2  5 x + 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
Simplifying complex fractions
Sometimes the numerator or the denominator of an algebraic fraction contains another
fraction. For example,
Simplify
3x  1
x
x2
This can be simplified by multiplying the numerator and the denominator by x.
3 x2  1
3 x  1x
=
2
x3
x
Simplify
1 2
a
a
a3
Multiply the numerator and the denominator by 3a:
1  2a
3a  6
3a  6
= 2
=
2
a
a  3 3a  a
4a 2
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 and 3x. That is ___.
3x2
3  42
9 x  12
x x
= 2
4
1 3x 3 x  4 x
3(3 x + 4)
=
x(3 x + 4)
=
3
x
Adding and subtracting algebraic fractions
Contents
Simplifying algebraic fractions
Adding and subtracting algebraic fractions
Multiplying and dividing algebraic fractions
Improper fractions and polynomial division
Examination-style question
12 of 37
© Boardworks Ltd 2006
Adding and subtracting fractions
Before looking at the addition and subtraction of algebraic fractions, let’s recall the
method used for numerical fractions.
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.
This is the lowest common multiple (LCM) of their denominators.
Adding and subtracting fractions
The LCM of 6 and 4 is ___.
So we write,
12
9
5 3 10
10 + 9
+ =
6 4
12
19
=
12
7
=1 12
We apply the same method to add or subtract algebraic fractions.
Adding and subtracting fractions
Write
2 as a3single fraction in its lowest terms.
+
2
x
x
The LCM of x2 and x is ___.
x2
3xx
2 3 22 + 3
+ =
2
x
x
x2
Write
y x
as
a single fraction in its lowest terms.
3x y
The LCM of 3x and y is ___.
3xy
3xx2 2
y x yy22  3
 =
3x y
3 xy
Adding and subtracting fractions
Write
2 x as a1single fraction in its lowest terms.
+
x + 3 2x + 6
Start by factorizing where possible:
The LCM of x + 3 and 2(x + 3) is ______.
2x
1
+
x + 3 2( x + 3)
2(x + 3)
2x
1
4x
1
+
=
+
x + 3 2( x + 3) 2( x + 3) 2( x + 3)
4 x +1
=
2( x + 3)
Adding and subtracting fractions
Write
2asx +
8 fraction in its lowest terms.
a
single
3 2
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
3 x2  2 x  7
=
x2 + 5
Notice that this
becomes – 8.
Multiplying and dividing algebraic fractions
Contents
Simplifying algebraic fractions
Adding and subtracting algebraic fractions
Multiplying and dividing algebraic fractions
Improper fractions and polynomial division
Examination-style question
18 of 37
© Boardworks Ltd 2006
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:
1
3
3 12
×
8 2 21 7
Then multiply the numerators and multiply the denominators:
1 3
3
× =
2 7 14
Multiplying and dividing fractions
To divide by a fraction we multiply by its reciprocal.
What is
7 ? 14
÷
9 15
This is equivalent to
1
5
7 15 1 5
×
= ×
9 3 14 2 3 2
5
=
6
Multiplying and dividing algebraic
fractions
We can apply the same methods to the multiplication and division of algebraic fractions.
For example,
Simplify
3 x  12
2
×
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
=
3
x+2
Multiplying and dividing algebraic
fractions
Simplify
x2  4
3x
×
x + 2 2x  4
x2  4
3x
( x  2)( x  2)
3x
×
=
×
x + 2 2x  4
x+2
2( x  2)
3x
=
2
Simplify
5 15
÷
2p p
1
5 15
5
p
÷
=
×
=
2p p
2 p 15 6
3
Multiplying and dividing algebraic
fractions
Simplify
14
7
÷ 2
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)
Improper fractions and polynomial division
Contents
Simplifying algebraic fractions
Adding and subtracting algebraic fractions
Multiplying and dividing algebraic fractions
Improper fractions and polynomial division
Examination-style question
24 of 37
© Boardworks Ltd 2006
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 to a mixed number as follows:
can be converted
6
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.
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).
Writing improper fractions in
proper form
r( x )
We can think of the form
asq(being
x ) +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.
Writing improper fractions in
proper form
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,
Write
x+3
in the form A +
x 1
.
B
x 1
x + 3 x  1+ 4
=
x 1
x 1
x 1
4
=
+
x 1 x 1
4
=1+
x 1
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.
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.
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.)
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 + Bx2  3 B + Cx + D
Equate the coefficients:
x 3:
A=1
x 2:
B = –4
x:
constants:
C – 3A = 0
C=3
D – 3B = 5
 D + 12 = 5
 D = –7
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 x 2 + 5
3x  7
 x 4+ 2
2
x 3
x 3
Examination-style question
Contents
Simplifying algebraic fractions
Adding and subtracting algebraic fractions
Multiplying and dividing algebraic fractions
Improper fractions and polynomial division
Examination-style question
35 of 37
© Boardworks Ltd 2006
Examination-style question
Given that
show that
f ( x)  x 
3
9
+ 2
{ x  , x  1, x  2},
x2 x  x2
x2 + x  3
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)
Examination-style question
Now divide x3 – x2 – 5x + 6 by x – 2:
x2
x–2
+ x
–3
x3 – x2 – 5x + 6
x3 – 2x2
x2 – 5x
x2 – 2x
–3x
+6
–3x + 6
0
So,
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