# factorization of algebraic expression

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Factorization of an algebraic expressions

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Content Enrichment Program Tilak College of Education Affiliated to University of Pune Prepared By- Sujata Jain Guided By- Prof. Namita Sahare

### Introduction:- Factors of an algebraic expression:-:

Introduction :- Factors of an algebraic expression:- When an algebraic expression is expressed as the product of two or more expressions, then each expression in the product is called the factor of the given algebraic expression. For example x 2 -y 2 can be expressed as the product of ( x+y ) and (x-y). Hence, ( x+y )and (x-y) are the factors of (x 2 -y 2 ).

### Different methods of factorization of algebraic expressions- (a) By taking out the common factors::

Different methods of factorization of algebraic expressions- (a) By taking out the common factors: If an expression contains a common factor in each term, divide each term by this common factor. Write the quotient within the bracket and the common factor outside the bracket Ex.1. Factorise : 5x + 20y Solution: 5x +20y = 5(x+4y). Here 5 and (x+4y) are the factors of 5x+20y.

### (b) By grouping:

(b) By grouping In this method the expression is rearranged in groups such that each group contains a common factors. Ex. 1. Factorise : ax + by + bx + ay Solution: ax + by + bx + ay = ax + bx + ay + by = x (a + b) + y (a + b) = (a + b)(x + y)

### (c)Factors of difference of two squares::

(c)Factors of difference of two squares: We know that the product of ( a+b ) and (a-b) is a 2 -b 2 . Therefore , ( a+b ) and (a-b) are factors of a 2 -b 2 . We may write this as follows: a 2 - b 2 = (a + b) (a - b) Ex. 1. Factorise : 4p 2 - 9q 2 Solution: 4p 2 - 9q 2 = (2p) 2 - (3q) 2 = (2p + 3q) (2p - 3q)

### (d) Factors of sum and difference of two cubes:

(d) Factors of sum and difference of two cubes We know that ( a+b ) 3 = a 3 +3a 2 b+3ab 2 +b 3 ( a+b ) 3 -3a 2 b-3ab 2 = a 3 +b 3 ( a+b ) 3 -3ab( a+b ) = a 3 +b 3 ( a+b ) [( a+b ) 2 -3ab] = a 3 +b 3 ( a+b ) (a 2 –ab+b 2 )= a 3 +b 3 a 3 +b 3 = ( a+b ) (a 2 –ab+b 2 ) Similarly, a 3 -b 3 =(a-b) (a 2 -ab+b 2 ) Ex.1. Factorise : 16 a 3 + 54b 3 =2(8a 3 +27b 3 ) =2[(2a) 3 +(3b) 3 ] =2(2a+3b) (4 a 2 -6ab+9b 2 )

### (e)Factorization of quadratic trinomial of the form ax2 +bx+c :

(e)Factorization of quadratic trinomial of the form ax 2 + bx+c Find the product of ‘a’ and ‘c’ . Find two factors of the product ‘ac’ such that their algebraic sum is b. Suppose that two factors are p and q. Express the middle term ‘ bx ’ as px+qx . The given trinomial will be in the form ax 2 + px+qx+c . Factorise this expression by grouping.

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Ex. Factorise 2 x 2 -x-6 Here a= 2, b=(-1), c=(-6) ac= 2* (-6) = (-12) (-x)= (-4x)+3x (-4)*(+3)= (-)12 2 x 2 -4x+3x-6 2x(x-2)+3(x-2) =(x-2) (2x+3)

### (f) Factorisation of quadratic trinomial of the form ax2+bxy+cy2:

(f) Factorisation of quadratic trinomial of the form ax 2 +bxy+cy 2 The trinomial of the form ax 2 +bxy+cy 2 can be factorised by the method similar to the method as discussed in factorizing the trinomial of the form ax 2 +bx+c. Ex.1. Factorise : x 2 +7xy+10y 2 Solution: x 2 +7xy+10y 2 = x 2 +2xy + 5xy+10y 2 Since 2*5= 10 = (x+2y) (x+5y) and 2+5=7

### (g) Factors of algebraic expressions reducible to quadratic trinomial:

(g) Factors of algebraic expressions reducible to quadratic trinomial There are some algebraic expressions which are not quadratic but they can be expressed in the form ax 2 +bxy+c. Once expressed in this form, we can then factorise them by the method discussed earlier.

### PowerPoint Presentation:

Ex.1. Factorise x 4 -5x 2 y 2 +4y 4 Solution: x 4 -5x 2 y 2 +4y 4 Let x 2 =a and y 2 =b x 4 =a 2 , y 4 = b 2 and x 2 y 2 = ab x 4 -5x 2 y 2 +4y 4 = a 2 -5ab+4b 2 = (a-4b)(a-b) Resubstituting the values of a and b we get, x 4 -5x 2 y 2 +4y 4 = (x 2 -4y 2 )(x 2 -y 2 ) = [ (x) 2 -(2y) 2 ] [(x) 2 -(y) 2 ] =[(x+2y)(x-2y)] [( x+y )(x-y)] = (x+2y)(x-2y) ( x+y )(x-y)

### (h) Factorisaton of an expression of the form a3+b3+c3-3abc:

(h) Factorisaton of an expression of the form a 3 +b 3 +c 3 -3abc Ex. Factorise a 3 +b 3 +c 3 -3abc Solution: a 3 +b 3 +c 3 -3abc We know that a 3 +b 3 =( a+b ) 3 -3ab( a+b ) a 3 +b 3 +c 3 -3abc= a 3 +b 3 +c 3 -3abc =( a+b ) 3 -3ab( a+b )+ c 3 -3abc =[( a+b ) 3 + c 3 ]-3ab( a+b )-3abc =[( a+b )+c] [( a+b ) 2 -( a+b )c+ c 2 ]-3ab( a+b+c ) =( a+b+c ) [a 2 +2ab+ b 2 -ac-bc+ c 2 -3ab ] =( a+b+c ) (a 2 +b 2 +c 2 -ab-bc-ca )

### (i) Factors of an expression of the form a3+b3+c3:

( i ) Factors of an expression of the form a 3 +b 3 +c 3 We know that a 3 +b 3 +c 3 -3abc=( a+b+c )(a 2 +b 2 +c 2 -ab-bc-ca) Case I: If a+b+c = 0 then show that a 3 +b 3 +c 3 =3abc. Solution: We know that a 3 +b 3 +c 3 -3abc = ( a+b+c )(a 2 +b 2 +c 2 -ab-bc-ca) a 3 +b 3 +c 3 -3abc = 0 *(a 2 +b 2 +c 2 -ab-bc-ca) as ( a+b+c = 0 ) a 3 +b 3 +c 3 -3abc = 0 a 3 +b 3 +c 3 = 3abc

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Case II: If a+b+c = 0 and a 3 +b 3 +c 3 -3abc =0,then show that a=b=c. Solution: We know that a 3 +b 3 +c 3 -3abc = ( a+b+c ) ( a 2 +b 2 +c 2 -ab-bc-ca) 0 = ( a+b+c ) ( a 2 +b 2 +c 2 -ab-bc-ca ) …given 0 = a 2 +b 2 +c 2 -ab-bc-ca a 2 +b 2 +c 2 -ab-bc-ca = 0 ½[2 a 2 +2b 2 +2c 2 -2ab-2bc-2ca]= 0 [ a 2 -2ab+b 2 +b 2 -2bc+c 2 +c 2 -2ca+a 2] = 0 [(a-b) 2 +(b-c) 2 +(c-a) 2 ]= 0 Continued

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But if the sum of no negative terms is zero then each term must be zero. Thus (a-b) 2 = 0 , (b-c) 2 = 0 and (c-a) 2 = 0 a-b = 0 , b-c = 0 and c-a = 0 a = b , b = c and c = a Thus a= b=c

### Questions:- :

Questions:- What is factorization of algebraic expressions? What are the different methods to factorise algebraic expressions? Factorise : (I) 1/3a 2 -3b (II) x 3 -3x 2 +x-3 (III) 50 x 3 -8x (IV) m 3 -1/m 3 (V)5 a 2 -18a-8 (VI) a 2 +2ab-24 b 2 (VII) x 4 -8x 2 y 2 +12y 4 (VIII) a 3 +b 3 + 8c 3 -6abc (IX) x 3 -27 y 3 +125+45xy

### Reference:-:

Reference :- Algebra – standard IX Maharashtra State Board of Secondary and Higher Secondary Education, Pune. 