Download Year 11 Maths Methods

TECHNIQUES IN FACTORISATION
The process where brackets are inserted into an equation is referred to as factorisation.
Factorisation is the opposite process to expansion.
Expansion
( x + 3)( x − 5)
→
←
x 2 − 2 x − 15
Factorisation
METHOD:
Bring all terms to one side of the equation and simplify by applying one or more of the
following techniques:
(a)
Remove the Highest Common Factor.
(b)
If the polynomial expression consists of TWO terms (binomial expression),
factorise by using one of the following rules:
• The difference of two squares.
a 2 − b 2 = (a − b )(a + b )
• The sum or difference of two cubes.
(
(
a 3 + b 3 = (a + b ) a 2 − ab + b 2
a 3 − b 3 = (a − b ) a 2 + ab + b 2
(c)
)
)
If the polynomial expression consists of THREE terms (trinomial expression),
factorise by using one of the following rules:
• Rules for perfect squares.
a 2 + 2ab + b 2 = (a + b )
a 2 − 2ab + b 2 = (a − b )
•
•
•
•
2
2
The FOIL method (to write expressions as linear factors).
Completing the Square.
Write the equation as a quadratic expression by using substitution (Let A = method).
The factor theorem and long division.
Note: A quadratic trinomial is an expression of the form: ax 2 + bx + c .
(d)
If the polynomial expression consists of FOUR items, factorise using:
• The factor theorem and long division.
• Grouping two and two.
• Grouping three and one.
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Summer School – Unit 1 Maths Methods – Book 1
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NUMBER SYSTEMS
Algebraic expressions are usually factorised over the following number systems:
•
•
•
The rational number system ( Q )
The real number system ( R )
The integer number system ( Z )
THE RATIONAL NUMBER SYSTEM (Q)
When factorising over Q , the factors can only consist of rational numbers i.e. whole
numbers or numbers that may be written as fractions.
x x + 2 x + 5 x2 +1
For example: ,
,
,
are rational numbers. Whole numbers may also be
2 3 x − 6 2x3 − 5
2
classified as rational numbers, as they too can be expressed as a fraction. Eg. 2 =
1
Note: Irrational numbers are expressions that cannot be expressed as a fraction.
For example: e , π and numbers that display an infinite number of decimal places.
THE REAL NUMBER SYSTEM (R)
When factorising over R , the factors can consist of both rational and irrational numbers
(surds).
For example:
1
x
, 2 , 10 ,
are real numbers.
2
3
− 5 is not part of the real number system as the square root of a negative number cannot
be evaluated.
THE INTEGER NUMBER SYSTEM (Z)
The integer number system includes positive and negative whole numbers, as well as zero.
i.e. ... − 3, − 2, − 1, 0, 1, 2, 3...
 The School For Excellence 2016
Summer School – Unit 1 Maths Methods – Book 1
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HIGHEST COMMON FACTORS
The highest common factor (HCF) is the highest single term that can be taken outside a
bracket. The HCF includes the largest common numeric factor (number), and the highest
common power of each pronumeral (letter).
Common factors should be removed before applying any other technique in factorisation.
Not only does the removal of a common factor make the factorisation process simpler, it may
also help to reveal what further techniques in factorisation need to be applied.
METHOD:
Step 1: Identify the numeric factors and letters that are common to all terms.
Step 2: Write the common factor outside a set of brackets.
Step 3: Write the given equation inside the brackets and divide each term by the
highest common factor.
Step 4: Simplify.
QUESTION 6
Factorise 4a 2 + 6ab .
Solution
Identify the HCF:
The highest term that divides into both 4a 2 and 6ab is 2a .
Write the HCF outside a set of brackets:
2a (
)
Divide each term by the HCF and simplify:
 4a 2 6ab 
 = 2a (2a + 3b)
4a 2 + 6ab = 2a
+
2a 
 2a
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Summer School – Unit 1 Maths Methods – Book 1
Page 12
QUESTION 7
Factorise 10 xy − 5 xy 2 + 5 x 2 y .
Solution
The HCF is 5 xy .
 10 xy 5 xy 2 5 x 2 y 

∴10 xy − 5 xy 2 + 5 x 2 y = 5 xy
−
+
5 xy
5 xy 
 5 xy
= 5 xy (2 − y + x)
QUESTION 8
Factorise x +
1
.
x
Solution
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Summer School – Unit 1 Maths Methods – Book 1
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THE DIFFERENCE OF TWO SQUARES (DOTS)
a 2 − b2 =
(
a 2 + b2
)(
)
a 2 − b 2 = ( a + b )( a − b )
The Difference of Two Squares (DOTS) is used to factorise equations that consist of the
difference of two terms (binomial expressions), both of which are perfect squares.
To factorise these expressions:
Step 1: Remove the highest common factor.
Step 2: Take the square root of each entire term.
Step 3: Add and subtract each term.
Note:
The sum of two squares ( a 2 + b 2 ) cannot be factorised.
QUESTION 9
Factorise each of the following expressions:
(a)
a2 − b2
+
+
a 2 − b2 = a 2 ± b2
= a±b
= (a + b)(a − b)
(b)
4 x 2 − 16 y 2
4 x 2 − 16 y 2 = 4( x 2 − 4 y 2 )
= 4( x 2 ± 4 y 2 )
= 4( x ± 2 y )
= 4( x + 2 y )( x − 2 y )
(c)
16 − 7 y 2
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(d)
( x − 1) 2 − 4
(e)
(2 x − 1) 2 − ( x + 1) 2
(f)
4( x − 2 y ) 2 − 9(2 x + y ) 2
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Summer School – Unit 1 Maths Methods – Book 1
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THE SUM AND DIFFERENCE OF TWO CUBES
The sum or difference of cubic binomials (two terms that are cubed) may be factorised using
the following rules:
The identity for the sum of two cubes is: a 3 + b 3 = (a + b)(a 2 − ab + b 2 )
The identity for the difference of two cubes is: a 3 − b 3 = (a − b)(a 2 + ab + b 2 )
Note:
•
•
a represents the cube root of the first term.
b represents the cube root of the second term.
•
•
The sign in between the two terms determines which formula is to be applied.
The quadratic term a 2 ± ab + b 2 cannot be factorised further.
(
)
To factorise these expressions:
Step 1: Remove the highest common factor.
Step 2: State the expressions for a and b (take the cube root of each given term).
Step 3: Substitute the expressions for a and b into the appropriate rule.
QUESTION 10
Factorise the following expressions:
(a)
8 x 3 + 27
This expression can be written in the form of the sum of two cubes:
As a 3 + b 3 = (a + b)(a 2 − ab + b 2 )
8 x 3 + 27 = (2 x) 3 + (3) 3
= (2 x + 3)[(2 x) 2 − (2 x)(3) + (3) 2 ]
= (2 x + 3)(4 x 2 − 6 x + 9)
(b)
a 3 − 125b 3
This expression can be written in the form of the difference of two cubes:
As a 3 − b 3 = (a − b)(a 2 + ab + b 2 )
a 3 − 125b 3 = a 3 − (5b) 3
= (a − 5b)[( a) 2 + ( a )(5b) + (5b) 2 ]
= (a − 5b)(a 2 + 5ab + 25b 2 )
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(c)
2( x − y ) 3 + 250
(d)
( x + y) 3 − ( x − y) 3
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QUADRATIC TRINOMIALS
A quadratic trinomial is an expression of the form: ax 2 + bx + c .
There are a variety of methods that may be used to factorise trinomials, including:
•
•
•
Rules for perfect squares.
Producing two linear factors by trial and error.
Completing the Square.
PERFECT SQUARES
If the given equation is a quadratic trinomial (a quadratic equation consisting of three terms),
where one term is two times the product of the square root of the other two terms, then the
equation may be factorised using the following rules:
a 2 + 2ab + b 2 = (a + b )
a 2 − 2ab + b 2 = (a − b )
2
2
Note that the sign in front of the term containing the product (2ab) determines which
formula is to be applied.
METHOD:
Step 1: Arrange terms in order of the power of the variable (usually x ).
Step 2: Take the square root of the first term.
Step 3: Carry down the sign in front of the coefficient of the middle term.
Step 4: Take the square root of the third term (the constant).
Step 5: Arrange the answer in the following format:
(square root of 1st term
sign in front of the 2nd term
square root of 3rd term)2
x 2 + 6x + 9
x2
(x
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9
+
3)
2
= ( x + 3)
2
Summer School – Unit 1 Maths Methods – Book 1
Page 18
QUESTION 11
Factorise the following expressions:
(a)
x 2 + 14 x + 49
(b)
25 x 2 + 20 xy + 4 y 2
(c)
9 x 2 − 30 xy + 25 y 2
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Summer School – Unit 1 Maths Methods – Book 1
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FACTORISING QUADRATIC TRINOMIALS
USING THE FOIL TECHNIQUE
Many quadratic trinomials (quadratic equations consisting of three terms) can be factorised
to produce two linear factors. This technique involves the construction of two pairs of
brackets, and inserting the appropriate factors by using FOIL in reverse.
METHOD:
Step 1: Bring all terms to one side of the equation and collect like terms.
Step 2: Arrange terms in order of the power on x .
Step 3: Write down the factors of the term involving x 2 .
Step 4: Write down the factors of the term that is independent of x (the constant).
Step 5: Choose factors in such a way that the sum of the products of the
(first term x last term) and the (outside term x inside term) is equal to the term
involving x .
For Example: Factorise x 2 + 2 xy + y 2 .
x 2 + 2 xy + y 2
=
(x
)( x
)
The product of the first two terms represents the term
involving x 2 .
x 2 + 2 xy + y 2
=
(x +
y )( x +
y)
The product of the second two terms represents the term that
is independent of x .
Factors must be chosen in such a way that the sum of the products of the
(first term x last term) and the (outside term x inside term) is equal to the term involving x .
( First
Outside)( Inside
Last )
This may involve some trial and error! Use the signs in front of the terms in the quadratic
equations to help in this process.
Note: There are four types of quadratic trinomials each characterised by different patterns
in signs.
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Summer School – Unit 1 Maths Methods – Book 1
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TYPE 1: TWO PLUS SIGNS
For Example: x 2 + 4 x + 3
1. This plus sign tells you that the
signs are the same in the two
brackets and that you ADD the
factors to find the middle term.
2. This plus sign tells
you that the signs
are positive.
( 3 × 1 = 3 and 3 + 1 = 4)
x 2 + 4x + 3
Therefore: x 2 + 4 x + 3 = ( x + 3)( x + 1)
TYPE 2: ONE MINUS AND ONE PLUS SIGN
For example: x 2 − 6 x + 5
2. This sign tells
you that the signs
are negative.
1. This plus sign tells you that the
signs are the same in the two
brackets and that you ADD the
factors to find the middle term.
( 5 × 1 = 5 and 5 + 1 = 6)
x 2 − 6x + 5
Therefore: x 2 − 6 x + 5 = ( x − 5)( x − 1)
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TYPE 3: TWO MINUS SIGNS
For Example: x 2 − 6 x − 7
2. This minus sign tells
you that the biggest factor is
negative.
1. This negative sign tells you that
the signs in the brackets are
different and that you subtract the
factors to find the middle term.
( 7 × 1 = 7 and 7 − 1 = 6)
x 2 − 6x − 7
Therefore: x 2 − 6 x − 7 = ( x − 7)( x + 1)
TYPE 4: ONE PLUS AND ONE MINUS SIGN
For Example: x 2 + x − 6
2. This plus sign tells
you that the biggest
factor is positive.
1. This negative sign tells you that
the signs in the brackets are
different and that you subtract the
factors to find the middle term.
(2 × 3 = 6 and 3 − 2 = 1)
x2 + x − 6
Therefore: x 2 + x − 6 = ( x + 3)( x − 2)
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QUESTION 12
Factorise the following expressions.
(a)
p 2 − 14 p − 15
p 2 − 14 p − 15 = ( p − 15)( p + 1)
(b)
x 2 − 3 x − 10
(c)
4 x 2 − 28 x + 49
4 x 2 − 28 x + 49 = (2 x − 7)(2 x − 7) = (2 x − 7) 2
(d)
12 + 11t − 15t 2
(e)
x 6 − 28 x 3 + 27
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(f)
6(2b − 1) 2 − 14(2b − 1) + 4
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Page 24