Factorisation

1. FACTORISATIONFACTORISATION
KONDETI YASHWANT
Class – VIII “A”

2. PREVEIW
• 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. Factorisation by regrouping terms
• 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. 5m2
n – 15mn2
5mn ( m -3n)
5 x m x n x(m-3n)

8. Factorise
a2
bc + ab2
c + abc2
Take the common factors
abc ( a + b +c )
a x b x c x (a + b + c)

9. Factorise
15pq + 15 + 9q + 25p
• 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
• 49p2
– 36
(7p)2
– (6)2
(7p+6) (7p-6)
•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
− − ÷ +
− + −
+
− + −
+
− +
+
−