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Contents
Simple
Algebraic Expressions
Evaluation of Algebraic Expression
Rules
of Algebra
Algebraic
Fractions
Quiz
1
Expansion:
The Foil Method
Algebraic
Identities
Basic
Factorisation
Factorisation of
Quadratic Polynomials
Factorisation
by Grouping
Quiz
2
Long
Division
Polynomial
Identities
Quiz
3
Algebra
Main Page
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Factorisation
of Quadratic Polynomials 
Example:
Factorise
x2 -5x +6 The
first term is x, which is the product of x and x. Therefore, the first term in
the bracket must be x, i.e., x2
- 5x + 6= (x ) (x ) The
last term is 6. The possible factors of 6 are thus  1
and 6
or 3
and 2.
Thus we have the following choices:
(x+
1) (x+6)
(x - 1) (x- 6)
(x+ 3) (x+2)
(x - 3) (x- 2)
The only pair of factors which gives -5x as the middle term
is (x-3) (x-2). Thus this is the answer.
The method below will help you to obtain the factors more
easily.
This is done by writing down the factors of the first term
and the last term in two columns. Then cross-multiply the factors and write
the products in the products in the third column. Add the products of the
third column. If the results is the same as the second term of the original
expression, then the factors will be those in the first two rows.
Factorise 3x2 -17x + 20
The factors of 3 x2 -17 x + 20 are 3x and x,
while those of 20 are 1
and 20
or 4
and 5
or 2
and 10.
By trial and error,
| First Trial |
Second Trial |
Third Trial |
 |
 |
 |
 3 x2 -17 x + 20
= (3x - 5) (x - 4)
With sufficient practices, the possible combination and trials can be done
mentally and many steps in the working may be omitted.
Example 2:
6x2
+33 xy +36 y2
= 3(2x2 +11 xy +12y2)
= 3(2x + 3) (x + 4)
10a2 +5ab -15b2
= 5(2a - b) (a +3b)
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