This document outlines the key topics and methods for factorizing algebraic expressions covered in a maths project. It includes factorizing natural numbers and algebraic expressions, using the method of common factors, factorizing by regrouping terms, solving sums using factorizing expressions and identities, factors of the form (x+a)(x+b), and dividing algebraic expressions including monomials and polynomials. Examples are provided for each method to demonstrate how to factorize different algebraic expressions step-by-step.

1. MATHS PROJECT
WORK

2. PREVIEW
PART1 :FACTORS OF NATURAL NUMBERS & AGEBRAIC EXPNS
PART2 :METHOD OF COMMON FACTORS
PART3: FACTORISATION BY REGROUPING TERMS
PART4 :SOLVING SUMS ON FACTORIZING EXPNS
PART5 :SOLVING SUMS ON FACTORISATION USING IDEN
PART6 :FACTORS OF THE FORM (x+a)(x+b)
PART7 :DIVISION OF ALGEBRAIC EXPNS

3. INTRODUCTION
Factors of natural numbers
e.g. 90 = 2 x 3 x 3 x 5
Factors of algebraic expressions
Terms are formed as product of factors.
e.g. 5xy = 5 x x x y
10x(x+2)(y+3)=2x5xxx(x+2)x(y+3)

4. METHOD OF COMMON FACTORS
We write each term as a product of irreducible
factors. Then take out the common factors of the
terms and write the remaining factors to get the
desired factor form.
E.g. 5ab+10a
(5xaxb)+(10xa)
(5axb) + (5ax2)
5a( b+2 ) (desired factor form)

5. Regrouping is re-arranging the terms with common terms.
e.g. 1) a2
+ ab + 8a + 8b
a(a+b) + 8(a+b)
(a+b) (a+8)
2) 15ab – 6a + 5b – 2
3a(5b-2) + 1(5b-2)
(3a+1) (5b-2)

6. LET US SOLVE SOME
EXAMPLES

7. EXAMPLE NO.1
1. 5m2
n – 15mn2
=5mn ( m -3n)
=5 x m x n x(m-3n)

8. FACTORISE:-
Take the common factors:-
=abc ( a + b +c )
=a x b x c x (a + b + c)

9. EXAMPLE 2:-
2. Factorise
15pq + 15 + 9q + 25p
ANS.)Regrouping like terms to find common factors
= 15pq + 9q + 25p +15
=3q ( 5p + 3 ) + 5 ( 5p + 3 )
=(3q + 5) (5p + 3)

10. FACTORISATION USING IDENTITIES
Observe the expression.
If it has a form that fits the right hand side of one of the identities ,
then the expression corresponding to the left hand side of the identity
gives the desired factorisation.
(a+b)2
= a2
+2ab+b2
(a-b)2
= a2
-2ab+b2
(a2
-b2
) = (a+b)(a-b)

11. FEW EXAMPLES:-
• 1) p2
+ 8p + 16
It is in the form of the identity a2
+2ab+b2
therefore
p2
+ 2 (p) (4) + 42
Since a2
+ 2ab +b2
= (a+b)2
By comparison
p2
+ 8p + 16 = ( p + 4)2
( required factorisation )

12. FACTORISE:-
1)49p2
– 36
=(7p)2
– (6)2
=(7p+6) (7p-6)
2) a2
– 2ab +b2
– c2
=(a-b)2
– c2
=(a-b-c) (a-b+c)

13. FACTORS OF THE FORM(x+a)(x+b)
In general , for factorising an algebraic expression of the
type x2
+px+q , we find two factors a and b of q (i.e. the
constant term) such that
ab = q
a+b = p

14. 4 2 2 4
2 2 2 2 2 2
2 2 2
2
( ) 2 ( )
( )
a a b b
a a b b
a b
− +
− +
+

15. 2
2
2
2
4 8 4
4( 2 1)
4( 1)
4[ ( 1) 1( 1)]
4( 1)( 1)
4( 1)
x x
x x
x x x
x x x
x x
x
− +
− +
− − +
− − −
− −
−

16. 2
2
6 8
4 2 8
( 4) 2( 4)
( 2)( 4)
p p
p p p
p p p
p p
+ +
+ + +
+ + +
+ +

17. 2
2
10 21
7 3 21
( 7) 3( 7)
( 7)( 3)
a a
a a a
a a a
a a
− +
− − +
− − −
− −

18. DIVISION OF ALGEBRAIC
EXPRESSIONS
Division of monomial by another monomial
Division of polynomial by a monomial
Division of polynomial by polynomial

19. Division of monomial by another
monomial
3
6 2x x÷
Now let us write the irreducible factor forms
3
3
2
3 2
6 2 3
2 2
6 2 (3 )
(2 ) (3 )
6 2 3
x x x x
x x
x x x x
x x
x x x
= ×× × ×
= ×
∴ = × × × ×
= ×
∴ ÷ =

20. Division of polynomial by a monomial
2
(5 6 ) 3
(5 6)
3
(5 6)
3
x x x
x x
x
x
− ÷
−
×
−

21. 3 2
2
2
( 2 3 ) 2
( 2 3)
2
( 2 3)
2
x x x x
x x x
x
x x
+ + ÷
+ +
×
+ +

22. Division of polynomial by polynomial
2
2
( 7 10) ( 5)
( 5 2 10)
( 5)
[ ( 5) 2( 5)]
( 5)
( 5)( 2)
( 5)
( 2)
y y y
y y y
y
y y y
y
y y
y
y
+ + ÷ +
+ + +
+
+ + +
+
+ +
+
+

23. 2
2
( 14 32) ( 2)
( 16 2 32)
( 2)
[ ( 16) 2( 16)]
( 2)
( 16)( 2)
( 2)
( 16)
m m m
m m m
m
m m m
m
m m
m
m
− − ÷ +
− + −
+
− + −
+
− +
+
−