FACTORISATION : FACTORISATION CLASS - 8
PREVEIW : PREVEIW PART1 FACTORS OF NATURAL NOS N ALGEBRAIC 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
DOUBT s??????
Introduction : 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)
Method of common factors : 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)
Factorisation by regrouping terms : 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)
Let us solve some examples : Let us solve some examples
5m2n – 15mn2 5mn ( m -3n)5 x m x n x(m-3n) : 5m2n – 15mn2 5mn ( m -3n)5 x m x n x(m-3n)
Factorise a2bc + ab2c + abc2 : Factorise a2bc + ab2c + abc2 Take the common factors
abc ( a + b +c )
a x b x c x (a + b + c)
Factorise 15pq + 15 + 9q + 25p : 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)
Factorisation using identities : 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)
Few examples : 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 )
Slide 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)
Factors of the form (x+a)(x+b) : 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
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Division of algebraic expressions : Division of algebraic expressions Division of monomial by another monomial
Division of polynomial by a monomial
Division of polynomial by polynomial
Division of monomial by another monomial : Division of monomial by another monomial Now let us write the irreducible factor forms
Division of polynomial by a monomial : Division of polynomial by a monomial
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Division of polynomial by polynomial : Division of polynomial by polynomial
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Slide 24 : ANY DOUBTS?
Slide 25 : THANK YOU !