SPECIAL PRODUCTS AND FACTORIZATION

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1 MODULE - Specil Products nd Fctoriztion 4 SPECIAL PRODUCTS AND FACTORIZATION In n erlier lesson you hve lernt multipliction of lgebric epressions, prticulrly polynomils. In the study of lgebr, we come cross certin products which occur very frequently. By becoming fmilir with them, lot of time nd lbour cn be sved s in those products, multipliction is performed without ctully writing down ll the steps. For emple, products, such s 08 08, 97 97, 04 96, , cn be esily clculted if you know the products ( b), ( b), ( b) ( b), ( b) respectively. Such products re clled specil products. Fctoriztion is process of finding the fctors of certin given products such s b, 8b, etc. We will consider fctoring only those polynomils in which coefficients re integers. In this lesson, you will lern bout certin specil products nd fctoriztion of certin polynomils. Besides, you will lern bout finding HCF nd LCM of polynomils by fctoriztion. In the end you will be mde fmilir with rtionl lgebric epressions nd to perform fundmentl opertions on rtionl epressions. OBJECTIVES After studying this lesson, you will be ble to write formule for specil products ( ± b), ( b) ( b), ( ) ( b), ( b) ( b b ), ( b) ( b b ), ( ± b) nd ( b) (c d); clculte squres nd cubes of numbers using formule; fctorise given polynomils including epressions of the forms b, ± b ; fctorise polynomils of the form b c ( 0) by splitting the middle term; determine HCF nd LCM of polynomils by fctoriztion; 00 Mthemtics Secondry Course

2 Specil Products nd Fctoriztion MODULE - cite emples of rtionl epressions in one nd two vribles; perform four fundmentl opertions on rtionl epressions. EXPECTED BACKGROUND KNOWLEDGE Number system nd four fundmentl opertions Lws of eponents ic epressions Four fundmentl opertions on polynomils HCF nd LCM of numbers Elementry concepts of geometry nd mensurtion lernt t primry nd upper primry levels. 4. SPECIAL PRODUCTS Here, we consider some speicl products which occur very frequently in lgebr. () Let us find ( b) ( b) ( b) ( b) ( b) b ( b) [Distributive lw] b b b b b Geometricl verifiction Concentrte on the figure, given here, on the right (i) ( b) Are of squre ABCD Are of squre AEFG re of rectngle EBIF re of rectngle DGFH re of squre CHFI b b b b b b G D H b C b b F I b A E b B b Thus, ( b) b b Mthemtics Secondry Course 0

3 MODULE - Specil Products nd Fctoriztion () Let us find ( b) ( b) ( b) ( b) [Distributive lw] ( b) b ( b) b b b b b Method : Using ( b) We know tht b ( b) ( b) [ ( b)] () ( b) ( b) b b Geometricl verifiction Concentrte on the figure, given here, on the right ( b) Are of squre PQRS Are of squre STVX X b P b S [re of rectngle RTVW re of rectngle PUVX re of squre QUVW] b( b) b ( b) (b b b ) W Q b R b b b b b b V b b U b( b) b T Thus, ( b) b b Deductions: We hve ( b) b b...() ( b) b b...() () () gives ( b) ( b) ( b ) () () gives ( b) ( b) 4b 0 Mthemtics Secondry Course

4 Specil Products nd Fctoriztion MODULE - () Now we find the product ( b) ( b) ( b) ( b) ( b) b ( b) [Distributive lw] b b b b A b D J b b Geometricl verifiction I H Observe the figure, given here, on the right ( b) ( b) Are of Rectngle ABCD Are of Rectngle AEFD b re of rectngle EBCF Are of Rectngle AEFD E b b F b G Are of Rectngle FGHI B b C [Are of Rectngle AEFD Are of rectngle FGHI Are of squre DIHJ] Are of squre DIHJ Are of squre AEGJ re of squre DIHJ b Thus, ( b) ( b) b The process of multiplying the sum of two numbers by their difference is very useful in rithmetic. For emple, (60 4) (60 4) (4) We, now find the product ( ) ( b) ( ) ( b) ( b) ( b) [Distributive lw] b b ( b) b Thus, Deductions: ( ) ( b) ( b) b (i) ( ) ( b) ( b) b (ii) ( ) ( b) (b ) b Mthemtics Secondry Course 0

5 MODULE - Specil Products nd Fctoriztion Students re dvised to verify these results. (5) Let us, now, find the product ( b) (c d) ( b) (c d) (c d) b (c d) c d bc bd c (d bc) bd Thus, ( b) (c d) c (d bc) bd Deductions: (i) ( b) (c d) c (d bc) bd (ii) ( b) (c d) c (bc d) bd Students should verify these results. Let us, now, consider some emples bsed on the specil products mentioned bove. Emple 4.: Find the following products: (i) ( b) (ii) 6b (iii) ( y) ( y) (iv) ( 9) ( ) (v) ( 5) ( 7) (vi) (5 8) (5 6) (vii) (7 ) (7 ) (viii) ( 5) ( 4) Solution: (i) Here, in plce of, we hve nd in plce of b, we hve b. ( b) () () (b) (b) 4 b 9b (ii) Using specil product (), we get 6b ( 6b) ( 6b) 9 4 8b 6b (iii) ( y) ( y) () y [using speicl product ()] 9 y (iv) ( 9) ( ) (9 ) 9 [using speicl product (4)] 04 Mthemtics Secondry Course

6 Specil Products nd Fctoriztion MODULE - 7 (v) ( 5) ( 7) (5 7) (vi) (5 8) (5 6) (5) (8 6) (5) (vii) (7 ) (7 ) (7) ( ) (7) () () (viii) ( 5) ( 4) ( ) ( 4 5 ) Numericl clcultions cn be performed more conveniently with the help of specil products, often clled lgebric formule. Let us consider the following emple. Emple 4.: Using specil products, clculte ech of the following: (i) 0 0 (ii) (iii) 68 7 (iv) 07 0 (v) (vi) Solution: (i) (00 ) (ii) (00 ) (iii) 68 7 (70 ) (70 ) (iv) 07 0 (00 7) (00) 00 (7 ) Mthemtics Secondry Course 05

7 MODULE - Specil Products nd Fctoriztion (v) (50 6) (50 ) 50 (6 ) (vi) (00 6) (00 ) 00 (6 ) CHECK YOUR PROGRESS 4.. Find ech of the following products: (i) (5 y) (ii) ( ) (iii) (b cd) (iv) ( 5y) (v) z (vi) (vii) ( 5) ( 5) (viii) (y ) (y ) 4 (i) 4 () (i) ( y) ( y) (ii) (7 5y) ( y). Simplify: (i) ( 5) ( 5) (ii) ( ) ( ) (iii) ( by) ( by) (iv) (p 8q ) (p 8q ). Using specil products, clculte ech of the following: (i) 0 0 (ii) (iii) (iv) (v) (vi) 57 4 (vii) (viii) (i) () 77 7 (i) (ii) Mthemtics Secondry Course

8 Specil Products nd Fctoriztion MODULE - 4. SOME OTHER SPECIAL PRODUCTS (6) Consider the binomil ( b). Let us find its cube. ( b) ( b) ( b) ( b) ( b b ) [using lws of eponents) ( b b ) b ( b b ) [Distributive lws) b b b b b b b b b( b) b Thus, ( b) b( b) b (7) We now find the cube of ( b). ( b) ( b) ( b) ( b) ( b b ) [using lws of eponents) ( b b ) b ( b b ) [Distributive lws) b b b b b b b b b( b) b Thus, ( b) b( b) b Note: You my lso get the sme result on replcing b by b in ( b) b( b) b (8) ( b)( b b ) ( b b ) b( b b ) [Distributive lw] b b b b b b Thus, ( b)( b b ) b (9) ( b)( b b ) ( b b ) b( b b ) [Distributive lw] b b b b b b Thus, ( b)( b b ) b Let us, now, consider some emples bsed on the bove mentioned specil products: Mthemtics Secondry Course 07

9 MODULE - Specil Products nd Fctoriztion Emple 4.: Find ech of the following products: (i) (7 9y) (ii) (p yz) (iii) ( 4y ) (iv) ( b ) 5 (v) b 4 (vi) c Solution: (i) (7 9y) (7) (7) (9y) (7 9y) (9y) 4 89 y (7 9y) 79y 4 y 70y 79y (ii) (p yz) (p) (p) (yz) (p yz) (yz) p pyz (p yz) y z p p yz py z y z (iii) ( 4y ) (4y ) ( 4y ) (4y ) y ( 4y ) 64y 6 y 48y 4 64y 6 (iv) ( b ) ( ) ( )(b ) ( b ) (b ) 5 (v) b b ( b ) 7b b 54 b 4 7b b b b b b b b b b 7 4 (vi) c () () c c c c c c c c 7 c 08 Mthemtics Secondry Course

10 Specil Products nd Fctoriztion Emple 4.4: Using specil products, find the cube of ech of the following: (i) 9 (ii) 0 (iii) 54 (iv) 47 Solution: (i) 9 ( 0 ) 0 0 (0 ) (0 ) MODULE - (ii) 0 ( 00 ) (00 ) (iii) 54 ( 50 4) (50 4) (50 4) (iv) 47 ( 50 ) (50 ) (50 ) Emple 4.5: Without ctul multipliction, find ech of the following products: (i) ( b) (4 6b 9b ) (ii) ( b) (9 6b 4b ) Solution: (i) ( b) (4 6b 9b ) ( b) [() () (b) (b) ] () (b) 8 7b (ii) ( b) (9 6b 4b ) ( b) [() () (b) (b) ] Mthemtics Secondry Course 09

11 MODULE - Specil Products nd Fctoriztion () (b) Emple 4.6: Simplify: 7 8b (i) ( y) ( y) ( y) ( y) ( y) ( y) (ii) ( b) ( b) (b ) ( b) (b ) Solution: (i) Put y nd y b The given epression becomes b b b ( b) ( y y) (6) (ii) 6 Put b nd b y so tht b y The given epression becomes y ( y) y ( y) ( b) b b b Emple 4.7: Simplify: (i) (ii) Solution: The given epression cn be written s Let 857 nd 57 b, then the epression becomes b b b ( b)( b b ) b b b 0 Mthemtics Secondry Course

12 Specil Products nd Fctoriztion (ii) The given epression cn be written s ( )( ) MODULE - CHECK YOUR PROGRESS 4.. Write the epnsion of ech of the following: b (i) ( 4y) (ii) (p qr) (iii) (iv) b (v) b (vi) b. Using specil products, find the cube of ech of the following: (i) 8 (ii) (iii) 8 (iv) (v) 5 (vi) 48 (vii) 7 (viii) 69 (i) 97 () 99. Without ctul multipliction, find ech of the following products: (i) ( y) (4 y y ) (ii) ( ) ( 4) (iii) ( ) ( ( ) (iv) (y z ) (4y 6yz 9z 4 ) (v) (4 y) (6 y 9y ) (vi) y 9 y y Find the vlue of: (i) 8b if b 0 nd b 5 [Hint: ( b) 8b 6b ( b) 8b ( b) 6b ( b)] (ii) y when y 5 nd y 66 y Mthemtics Secondry Course

13 MODULE - Specil Products nd Fctoriztion 5. Find the vlue of 64 5z if (i) 4 5z 6 nd z (ii) 4 5z 5 nd z 6 6. Simplify: (i) ( 5) ( 5) (ii) (7 5y) (7 5y) 0y (7 5y) (7 5y) [Hint put 7 5y nd 7 5y b so tht b 0y] (iii) ( y) (9 6y 4y ) ( y) (4 6y 9y ) (iv) ( 5) (4 0 5) (5 ) (5 5 ) 7. Simplify: (i) (ii) FACTORIZATION OF POLYNOMIALS Recll tht from 4, we sy tht nd 4 re fctors of the product. Similrly, in lgebr, since ( y) ( y) y, we sy tht ( y) nd ( y) re fctors of the product ( y ). Fctoriztion of polynomil is process of writing the polynomil s product of two (or more) polynomils. Ech polynomil in the product is clled fctor of the given polynomil. In fctoriztion, we shll restirct ourselves, unless otherwise stted, to finding fctors of the polynomils over integers, i.e. polynomils with integrl coefficients. In such cses, it is required tht the fctors, too, be polynomils over integers. Polynomils of the type y will not be considered s being fctorble into ( y)( y) these fctors re not polynomils over integers. becuse A polynomil will be sid to be completely fctored if none of its fctors cn be further epressed s product of two polynomils of lower degree nd if the integer coefficients hve no common fctor other thn or. Thus, complete fctoriztion of ( 4) is ( 4). On the other hnd the fctoriztion (4 ) (4 ) of (6 4 ) is not complete since the fctor (4 ) cn be further fctorised s ( ) ( ). Thus, complete fctoriztion of (6 4 ) is ( ) ( ) (4 ). In fctoriztion, we shll be mking full use of specil products lernt erlier in this lesson. Now, in fctoriztion of polynomils we tke vrious cses seprtely through emples. Mthemtics Secondry Course

14 Specil Products nd Fctoriztion () Fctoriztion by Distributive Property Emple 4.8: Fctorise: (i) 0 5 (ii) y y (iii) 5b ( y ) 6mn( y ) (iv) (b c) b(b c) Solution: (i) ( 5) [Since 5 is common to the two terms] Thus, 5 nd 5 re fctors of 0 5 (ii) In y y, note tht y is common (with gretest degree) in both the terms. y y y y y y (y ) MODULE - Therefore,,, y, y, y, y, y, y nd y re fctors of y y (iii) Note tht y is common in both the terms 5b ( y ) 6mn( y ) ( y ) (5b 6mn) (iv) (b c) b(b c) (b c) [(b c)] (b c) b (b c) [(b c) b] (b c) [b c b] () Fctoriztion Involving the Difference of Two Squres You know tht ( y) ( y) y. Therefore y nd y re fctors of y. Emple 4.9: Fctorise: (i) 9 6y (ii) 4 8y 4 (iii) 4 (b c) (iv) y 6y 9 Solution: (i) 9 6y () (4y) which is difference of two squres. ( 4y) ( 4y) (ii) 4 8y 4 ( ) (9y ) ( 9y ) ( 9y ) Note tht 9y () (y) is gin difference of the two squres. 4 8y 4 ( 9y ) [() (y) ] ( 9y ) ( y) ( y) Mthemtics Secondry Course

15 MODULE - Specil Products nd Fctoriztion (iii) 4 (b c) ( ) (b c) [ (b c)] [ (b c)] ( b c) ( b c) (iv) y 6y 9 (y 6y 9) [Note this step) () [(y) y () ] () (y ) () Fctoriztion of Perfect Squre Trinomil Emple 4.0 : Fctorise [ (y )] [ (y )] ( y ) ( y ) (i) 9 4y 6y (ii) Solution: (i) 9 4y 6y () () (4y) (4y) ( 4y) ( 4y) ( 4y) Thus, the two fctors of the given polynomil re identicl, ech being ( 4y). (ii) ( ) ( ) (4) (4) ( 4) ( 4) ( 4) Agin, the two fctors of the given polynomil re identicl, ech being ( 4). (4) Fctoriztion of Polynomil Reducible to the Difference of Two Squres Emple 4.: Fctorise (i) 4 4y 4 (ii) 4 Solution: (i) 4 4y 4 ( ) (y ) ( ) (y ) ( ) (y ) ( ) (y ) [Adding nd subtrcting ( ) (y )] ( y ) (y) ( y y) ( y y) 4 Mthemtics Secondry Course

16 Specil Products nd Fctoriztion MODULE - (ii) 4 ( ) () [Adding nd subtrcting ] ( ) () ( ) ( ) CHECK YOUR PROGRESS 4. Fctorise:. 0 y 5z. bc b c. 6p 5pq 7 p 4. (b c) b (c b) 5. (4 y) b (4 y) 6. ( y) y ( y) p 8. 56y 8 9. ( ) 9 0. ( bc) (b c) y. 49 4y y. m 4m b Find the vlue of n if (i) 6n 7 7 (ii) n (5) Fctoriztion of Perfect Cube Polynomils Emple 4.: Fctorise: (i) 6 y y 8y (ii) 6 4 y y 4 y 6 Solution: (i) 6 y y 8y () (y) (y) (y) ( y) Thus, the three fctors of the given polynomil re identicl, ech being y. (ii) Given polynomil is equl to ( ) y ( y ) (y ) ( y ) [( y) ( y)] [Since y ( y) ( y)] ( y) ( y) Mthemtics Secondry Course 5

17 MODULE - Specil Products nd Fctoriztion (6) Fctoriztion of Polynomils Involving Sum or Difference of Two Cubes In specil products you hve lernt tht ( y) ( y y ) y nd ( y) ( y y ) y Therefore, the fctors of y re y nd y y nd those of y re y nd y y Now, consider the following emple: Emple 4.: Fctorise (i) 64 7b (ii) 8 5y (iii) 8 ( y) 4 (iv) 4 Solution: (i) 64 7b (4) (b) (4 b) [(4) (4)(b) (b) ] (4 b) (6 b 9b ) (ii) 8 5y () (5y) ( 5y) [() ()(5y) (5y) ] ( 5y) (4 0y 5y ) (iii) 8 ( y) 4 [( y)] (7) [( y) 7] [ ( y) ( y) (7) 7 ] ( 4y 7) (4 6y 6y 4 8y 49) (iv) 4 4 ( 9 ) [Since 4 is common to the two terms] 4 [() ( ) ] 4 ( ) ( 6 ) 4 ( ) ( ) ( 6 ) [Since ( ) ( )] CHECK YOUR PROGRESS 4.4 Fctorise:. 6b. 4. y 48y 64y y 54y 7y 6 Mthemtics Secondry Course

18 Specil Products nd Fctoriztion y 60 y 50y 6. 64k 44k 08k y b b. ( b) 64c. 64 (y ) MODULE - (7) Fctorising Trinomils by Splitting the Middle Term You hve lernt tht ( ) ( b) ( b) b. ( b) b nd ( b) (c d) c (d bc) bd In generl, the epressions given here on the right re of the form A B C which cn be fctorised by multiplying the coefficient of in the first term with the lst term nd finding two such fctors of this product tht their sum is equl to the coefficient of in the second (middle) term. In other words, we re to determine two such fctors of AC so tht their sum is equl to B. The emple, given below, will clrify the process further. Emple 4.4:Fctorise: (i) (ii) 0y 4y (iii) 5 6 (iv) Solution: (i) Here, A, B nd C ; so AC (ii) Therefore we re to determine two fctors of whose sum is Obviously, (i.e. two fctors of AC i.e. re nd ) We write the polynomil s ( ) ( ) ( ) ( ) ( ) Here, AC 4y nd B 0y Two fctors of 4y whose sum is 0y re 4y nd 6y We write the given polynomil s 4y 6y 4 y ( 4y) 6y( 4y) ( 4y) ( 6y) Mthemtics Secondry Course 7

19 MODULE - Specil Products nd Fctoriztion (iii) Here, AC 5 ( 6) 0 nd B Two fctors of 0 whose sum is re 5 nd We write the given polynomil s ( ) ( ) ( ) (5 ) (iv) Here, AC ( ) 6 nd B Two fctors of 6 whose sum is ( ) re ( ) nd. We write the given polynomil s ( ) ( ) ( ) ( ) CHECK YOUR PROGRESS 4.5 Fctorise:. 4. 5y 54y y 4y y 0y y y (m ) ( m) m 09. ( b) ( b) 0. ( y) ( y)( y) ( y) Hint put b Hint: Put y nd y b 4.4 HCF AND LCM OF POLYNOMIALS () HCF of Polynomils You re lredy fmilir with the term HCF (Highest Common Fctor) of nturl numbers in rithmetic. It is the lrgest number which is fctor of ech of the given numbers. For instnce, the HCF of 8 nd is 4 since the common fctors of 8 nd re, nd 4 nd 4 is the lrgest i.e. highest mong them. On similr lines in lgebr, the Highest Common Fctor (HCF) of two or more given 8 Mthemtics Secondry Course

20 Specil Products nd Fctoriztion polynomils is the product of the polynomil(s) of highest degree nd gretest numericl coefficient ech of which is fctor of ech of the given polynomils. For emple, the HCF of 4( ) nd 6( ) is ( ). The HCF of monomils is found by multiplying the HCF of numericl coefficients of ech of the monomils nd the vrible(s) with highest power(s) common to ll the monomils. For emple, the HCF of monomils y, 8y 4 nd 4 y 5 is 6y since HCF of, 8 nd 4 is 6; nd the highest powers of vrible fctors common to the polynomils re nd y. Let us now consider some emples. Emple 4.5: Find the HCF of MODULE - (i) 4 y nd y (ii) ( ) ( ) nd ( ) ( ) Solution: (i) HCF of numericl coefficients 4 nd is. Since occurs s fctor t lest twice nd y t lest once in the given polynomils, therefore, their HCF is y i.e. y (ii) HCF of numericl coefficients nd is. In the given polynomils, ( ) occurs s fctor t lest twice nd ( ) t lest once. So the HCF of the given polynomils is ( ) ( ) i.e. ( ) ( ) In view of Emple 4.5 (ii), we cn sy tht to determine the HCF of polynomils, which cn be esily fctorised, we epress ech of the polynomils s the product of the fctors. Then the HCF of the given polynomils is the product of the HCF of numericl coefficients of ech of the polynomils nd fctor (s) with highest power(s) common to ll the polynomils. For further clrifiction, concentrte on the Emple 4.6 given below. Emple 4.6:Find the HCF of (i) 4 nd 4 4 (ii) nd Solution: (i) 4 ( ) ( ) 4 4 ( ) HCF of numericl coefficients HCF of other fctors ( ) Hence, the required HCF (ii) ( 4 ) 4 ( ) ( ) Mthemtics Secondry Course 9

21 MODULE - () LCM of Polynomils Specil Products nd Fctoriztion ( ) 6 ( 4) ( ) Required HCF ( ) [Since HCF of numericl coefficient is ) 6 Like HCF, you re lso fmilir with the LCM (Lowest Common Multiple or Lest Common Multiple) of nturl numbers in rithmetic. It is the smllest number which is multiple of ech of the given numbers. For instnce, the LCM of 8 nd is 4 since 4 is the smllest mong common multiples of 8 nd s given below: Multiples of 8: 8, 6, 4,, 40, 48, 56, 64, 7, 80,... Multiples of :, 4, 6, 48, 60, 7, 84, 96,... Common multiple of 8 nd : 4, 48, 7,... On similr lines in, the Lowest Common Multiple (LCM) of two or more polynomils is the product of the polynomil(s) of the lowest degree nd the smllest numericl coefficient which re multiples of the corresponding elements of ech of the given polynomils. For emple, the LCM of 4( ) nd 6( ) is ( ). The LCM of monomils is found by multiplying the LCM of numericl coefficients of ech of the monomils nd ll vrible fctors with highest powers. For emple, the LCM of y z nd 8 yz is 6 y z since the LCM of nd 8 is 6 nd highest powers vrible fctors, y nd z re, y nd z respectively. Let us, now, consider some emples to illustrte. Emple 4.7: Find the LCM of (i) 4 y nd y (ii) ( ) ( ) nd ( ) ( ) Solution: (i) LCM of numericl coefficient 4 nd is 4. Since highest power of is nd tht of y is y, the required LCM is 4 y (ii) Obviously LCM of numericl coefficients nd is. In the given polynomils, highest power of the fctor ( ) is ( ) nd tht of ( ) is ( ). LCM of the given polynomils ( ) ( ) ( ) ( ) 0 Mthemtics Secondry Course

22 Specil Products nd Fctoriztion In view of Emple 4.7 (ii), we cn sy tht to determine the LCM of polynomils, which cn be esily fctorised, we epress ech of the polynomils s the product of fctors. Then, the LCM of the given polynomils is the product of the LCM of the numericl coefficients nd ll other fctors with their highest powers which occur in fctoriztion of ny of the polynoils. For further clrifiction, we tke Emple 4.8 given below. MODULE - Emple 4.8:Find the LCM of (i) ( ) ( ) nd 5 6 (ii) 8( 7) nd ( 5 7 ) Solution: (i) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Also 5 6 ( ) ( ) LCM of numericl coefficients LCM of other fctors ( ) ( ) ( ) Hence, the LCM of given polynomils ( ) ( ) ( ) (ii) 8( 7) 8( ) ( 9) ( 5 7 ) ( 7) ( ) ( 9) LCM of numericl coefficient 8 nd 4 LCM of other fctors ( ) ( ) ( 9) ( 9) Hence, required LCM 4 ( ) ( ) ( 9) ( 9) CHECK YOUR PROGRESS 4.6. Find the HCF of the following polynomils: (i) 7 4 y nd y (ii) 48y 7 9 nd y 5 (iii) ( ) nd ( ) ( ) (iv) 4 4 nd (v) 8 ( ) nd 4 ( 8) (vi) ( ) ( 5) nd 0 5 (vii) ( 5) ( 4) nd ( 5) ( 4) (viii) nd 4 (i) y nd y () 6( ) nd 8( 4 ). Find the LCM of the following polynomils: (i) 5 y nd 5y (ii) 0 y nd 48 y 4 (iii) ( ) nd ( ) ( ) (iv) 4 4 nd (v) 8 ( ) nd 4 ( 8) (vi) ( ) ( 5) nd 0 5 (vii) ( 5) ( 4) nd ( 5) ( 4) (viii) nd 4 (i) y nd y () 6( ) nd 8( 4 ) Mthemtics Secondry Course

23 MODULE - Specil Products nd Fctoriztion 4.5 RATIONAL EXPRESSIONS You re lredy fmilir with integers nd rtionl numbers. Just s number, which cn be epressed in the form q p where p nd q ( 0) re integers, is clled rtionl number, n lgebric epression, which cn be epressed in the from Q P, where P nd Q (non-zero polynomils) re polynomils, is clled rtionl epression. Thus, ech of the epressions, 5, 5 b b 5 6, y y is rtionl epression in one or two vribles. : () The polynomil is rtionl epresion since it cn be written s nd you hve lernt tht the constnt in the denomintor is polynomil of degree zero. 7 () The polynomil 7 is rtionl epresion since it cn be written s where both 7 nd re polynomils of degree zero. () Obvioulsy rtionl epression need not be polynomil. For emple rtionl epression ( ) is not polynomil. On the contrry every polynomil is lso rtionl epression. None of the epressions,, b is rtionl epression. b b CHECK YOUR PROGRESS 4.7. Which of the following lgebric epressions re rtionl epressions? (i) 4 8 (ii) y Mthemtics Secondry Course

24 Specil Products nd Fctoriztion MODULE - (iii) 7 5 (iv) 6 (v) 00 (vi) b b (vii) y yz (y z) z (vii) 5 ( b). For ech of the following, cite two emples: (i) A rtionl epression is one vrible (ii) A rtionl epression is two vribles (iii) A rtionl epression whose numertor is binomil nd whose denomintor is trinomil (iv) A rtionl epression whose numertor is constnt nd whose denomintor is qudrtic polynomil (v) A rtionl epression in two vribles whose numertor is polynomil of degree nd whose denomintor is polynomil of degree 5 (vi) An lgebric epression which is not rtionl epression 4.6 OPERATIONS ON RATIONAL EXPRESSIONS Four fundmentl opertions on rtionl epressions re performed in ectly the sme wy s in cse of rtionl numbers. () Addition nd Subtrction of Rtionl Epressions For observing the nlogy between ddition of rtionl numbers nd tht of rtionl epressions, we tke the following emple. Note tht the nlogy will be true for subtrction, multipliction nd division of rtionl epressions lso. Emple 4.9:Find the sum: (i) 5 (ii) 6 8 Solution: (i) LCM of 6 nd Mthemtics Secondry Course

25 MODULE - Specil Products nd Fctoriztion (ii) ( )( ) ( )( ) ( )( ) LCM of ( ) nd ( ) 4 Emple 4.0: Subtrct Solution: from ( )( ) ( )( ) ( )( ) 4 ( ) ) 4 Note: Observe tht the sum nd difference of two rtionl epressions re lso rtionl epressions. Since the sum nd difference of two rtionl epressions re rtionl epressions, ( 0) nd ( 0) re both rtionl epressions s nd re both rtionl epressions. Similrly, ech of,,,, etc. is rtionl epression. These epresions crete interest s for given vlue of or, we cn determine vlues of,,, etc. nd in some cse vice vers lso. Let us concentrte on the following emple. Emple 4.:Find the vlue of 4 (i) if (ii) if (iii) if 9 4 (iv) if 4 Mthemtics Secondry Course

26 Specil Products nd Fctoriztion MODULE - Mthemtics Secondry Course 5 (v) 5 if Solution: (i) We hve () Hence, (ii) 4 ( ) ( ) So, 4 4 (iii) We hve ( ) 9

27 Specil Products nd Fctoriztion MODULE - Mthemtics Secondry Course 6 ( ) [since both nd re positive] 9 () ± (iv) We hve () 7 () 7 8 (v) We hve 5 () 5 5 ()

28 Specil Products nd Fctoriztion MODULE - CHECK YOUR PROGRESS 4.8. Find the sum of rtionl epressions: (i) nd (ii) nd nd (iii) ( ) (iv) 6 nd 5 ( ) 4 (v) nd (vi) nd (vii) nd (vii) nd. Subtrct (i) from 4 (ii) from (iii) from (iv) from (v) 4 from 4 from (vi) ( ) 9 from (vii) ( ) ( ) (vii) from 4. Find the vlue of (i) when (ii) when (iii) when (iv) when 5 (v) when 5 (vi) 8 when 5 7 (vii) when (viii) 7, 0 when > Mthemtics Secondry Course 7

29 MODULE - Specil Products nd Fctoriztion 4 (i) when 77 4 () 4, 0 4 when > 4 () Multipliction nd Division of Rtionl Epressions You know tht the product of two rtionl numbers, sy, 5 nd Similrly, the product of two rtionl epressions, sy, 7 7 is given s P R PR where P, Q, R, S (Q, S 0) re polynomils is given by. You my observe Q S QS tht the product of two rtionl epressions is gin rtionl epression. Emple 4.: Find the product: P Q nd R S (i) 5 5 (ii) (iii) Solution: ( 4) 5 5 (i) 5 ( 5 )( ) ( 5 )( ) (ii) ( )( ) ( )( ) (iii) 7 0 ( 4) 5 [Cncelling common fctor ( ) from numertor nd denomintor] 7 ( )( ) ( 4) ( 5) ( )( 5)( )( 4) ( 4) ( 5) 8 Mthemtics Secondry Course

30 Specil Products nd Fctoriztion ( )( ) ( 4) MODULE - [Cncelling common fctor ( 4) ( 5) from numertor nd denomintor] Note: The result (product) obtined fter cncelling the HCF from its numertor nd denomintor is clled the result (product) in lowest terms or in lowest form. You re lso fmilir with the division of rtionl number, sy, by rtionl number, sy, is given s where is the reciprocl of. Similrly, division of P R rtionl epression by non-zero rtionl epression is given by Q S where P, Q, R, S re polynomils nd R S is the reciprocl epression of S R. Emple 4.: Find the reciprocl of ech of the following rtionl epressions: P Q R S P Q S R (i) y (ii) y (iii) 5 8 Solution: (i) Reciprocl of is y y 5 5 y (ii) Reciprocl of y is 5 y y Emple 4.4: Divide: 8 (iii) Since 8, the reciprocl of 8 is 8 (i) by (ii) 5 by 4 5 nd epress the result in lowest form. 4 5 Mthemtics Secondry Course 9

31 MODULE - Solution: (i) (ii) 5 Specil Products nd Fctoriztion ( )( ) ( ) ( )( 4 5) ( 5)( 4 ( )( )( 5)( ) ( 5)( 5)( )( 5) ( )( ) ( 5)( 5) [Cncelling HCF ()(5)] 0 5 The result 0 5 is in lowest form. CHECK YOUR PROGRESS 4.9. Find the product nd epress the result in lowest terms: (i) (ii) 4 4 (iii) (iv) (v) (vi) (vii) (viii). Find the reciprocl of ech of the following rtionl epressions: 4 6 (i) (ii) 0 Mthemtics Secondry Course

32 Specil Products nd Fctoriztion (iii) 7 (iv) 4. Divide nd epress the result s rtionl epression in lowest terms: (i) (ii) MODULE - (iii) 9 4 (iv) (v) 4 5 (vi) 5 ( ) LET US SUM UP Specil products, given below, occur very frequently in lgebr: (i) ( y) y y (ii) ( y) y y (iii) ( y) ( y) y (v) ( b) (c d) c (d bc) bd (iv) ( ) ( b) ( b) b (vi) ( y) y( y) y (vii) ( y) y( y) y (viii) ( y) ( y y ) y (i) ( y) ( y y ) y Fctoriztion of polynomil is process of writing the polynomil s product of two (or more) polynomils. Ech polynomil in the product is clled fctor of the given polynomil. A polynomil is sid to be completely fctorised if it is epressed s product of fctors, which hve no fctor other thn itself, its negtive, or. Aprt from the fctoriztion bsed on the bove mentioned specil products, we cn fctorise polynomil by tking monomil fctor out which is common to some or ll of the terms of the polynomil using distributive lws. HCF of two or more given polynomils is the product of the polynomil of the highest degree nd gretest numericl coefficient ech of which is fctor of ech of the given polynomils. LCM of two or more given polynomils is the product of the polynomil of the lowest degree nd the smllest numericl coefficient which re multiples of corresponding elements of ech of the given polynomils. Mthemtics Secondry Course

33 MODULE - Specil Products nd Fctoriztion An lgebric epression, which cn be epressed in the form Q P where P nd Q re polynomils, Q being non-zero polynomil, is clled rtionl epression. Opertions on rtionl epressions re performed in the wy, they re performed in cse of rtionl numbers. Sum, Difference, Product nd Quotient of two rtionl epressions re lso rtionl epressions. Epressing rtionl epression into lowest terms mens cncelltion of common fctor, if ny, from the numertor nd denomintor of the rtionl eprssion. TERMINAL EXERCISE. Mrk tick ginst the correct lterntive: (i) If 0 0 5p, then p is equl to (A) 6 (B) 40 (C) 560 (D) 4000 (ii) ( ) ( ) is equl to (A) 4 (B) 4 4 (C) 7 (D) 7 4 (iii) ( b ) ( b ) is equl to (A) ( b ) (B) 4( b ) (C) 4( 4 b 4 ) (D) ( 4 b 4 ) (iv) If m, then m m m is equl to (A) 0 (B) 6 (C) 6 (D) (v) is equl to (A) 650 (B) 7 (C) (D) 4 (vi) 8m n is equl to: (A) (m n)(4m mn n ) (B) (m n)(4m mn n ) (C) (m n)(4m 4mn n ) (D) (m n)(4m 4mn n ) (vii) is equl to (A) 66 (B) 98 (C) 000 (D) 000 Mthemtics Secondry Course

34 Specil Products nd Fctoriztion (viii) The HCF of 6 5 b nd 90 b 4 is (A) 6 b (B) 8 b (C) 90 b 4 (D) 80 5 b 4 (i) The LCM of nd is (A) ( ) ( ) (B) ( ) ( ) (C) ( ) ( ) (D) ( ) ( ) () Which of the following is not rtionl epression? MODULE - (A) (B) 5 (C) 8 6 y (D). Find ech of the following products: (i) ( m n )( m n ) (ii) ( y )( y ) (iii) ( y) ( y) (iv) ( 5b)( 5b) (v) (5 y) ( 5 0y 4y ) (vi) ( 5y) (4 0y 5y ) 5 4 (vii) 4 5 (viii) (z )(z 5) (i) () (i) ( b 5) ( b 6) (ii) ( 7z) ( 5z). If b nd y b c, show tht ( c) ( c b) y 4. Find the vlue of 64 5z if 4 5z 6 nd z. 5. Fctorise: (i) 7 y 6 y 0 (ii) 5 b 4b 5 (iii) 6 4 b b 4 (iv) 4 8 b 6 b 6 (v) 4 y 4 (vi) (vii) 6 6 (viii) 7 (i) 7 y 6y () (i) 6 79y 6 (ii) b 6 6. Find the HCF of (i) 5 nd 4 7 Mthemtics Secondry Course

35 MODULE - Specil Products nd Fctoriztion (ii) 0( ) nd 50( ) 7. Find the LCM of (i) y nd y (ii) 4 y y 4 nd y y 8. Perform the indicted opertion: (i) (ii) (iii) (iv) ( ) Simpify: [Hint : 4 ; now combine net term nd so on] 0. If m nd n, find m n mn. ANSWERS TO CHECK YOUR PROGRESS 4.. (i) 5 0y y (ii) 6 9 (iii) b bcd c d (iv) 4 0y 5y (v) (vi) 9 z z (vii) 4 5 (viii) y (i) 4 Mthemtics Secondry Course

36 Specil Products nd Fctoriztion MODULE - () (i) y 6y (ii) 8y 5y. (i) 40 (ii) 6 8 (iii) ( b y ) (iv) p q. (i) 0404 (ii) 664 (iii) 476 (iv) (v) 684 (vi) 45 (vii) (viii) 4996 (i) 445 () 56 (i) 890 (ii) (i) 7 6 y 6y 64y (ii) p p qr pq r q r (iii) b b b (iv) 7 b b 7 b (v) b b b (vi) (i) 5 (ii) 78 (iii) 58 (iv) 67 (v) (vi) 059(vii) 579 (viii) 8509 (i) 966 () (i) 8 y (ii) 8 (iii) (iv) 8y 7z 6 (v) 64 7y (vi) 7 4. (i) 00 (ii) 5 5. (i) 566 (ii) b y 4 y b 6 y 4 8b 6. (i) 0 50 (ii) 000y (iii) 9 9y (iv) (i) 000 (ii) (y z). bc (c b). p(p 5q 9) 4. (b c) ( b) 5. (4 y) (8 y b) 6. ( y) ( y y ) 7. 5( 5p) ( 5p) 8. ( 6y 4 ) ( 4y ) ( y) ( y) 9. (5 ) ( ) 0. ( bc b c) ( bc b c) 9 y 6 Mthemtics Secondry Course 5

37 MODULE - Specil Products nd Fctoriztion. (5 6y ) (5 6y ). (7 y )(7 y ). (m 7) 4. ( ) 5. (6 5) 6. ( 4) 7. ( 4 7 ) ( )( ) 8. ( 6b 9b )( 6b 9b ) 9. ( )( ) 0. ( 5 4)( 5 4). (i) 40 (ii) ( 6b) ( 6b 6b ). ( 7) ( 7 49). ( 4y) 4. ( y) 5. ( 5y) 6. (4k ) 7. (9 ) ( ) 8. ( y ) ( y y 4 ) 9. ( b ) (4 6 b 9b 4 ) 0. (b ) (9b b b ). ( b 4c)(4 9b 6b 8c bc 6 c. (4 y )(6 8y 4 4y 4y ) 4.5. ( ) ( 8). ( 6y) ( 9y). ( ) ( ) 4. ( y)( y) 5. ( ) ( ) ( ) 6. ( 5y) ( y) 7. ( ) ( 7) 8. (y )(5y ) 9. ( ) ( ) 0. ( m) (m 9). ( b 6)( b 5). (9y 7)(5 y) 4.6. (i) y (ii) y 5 (iii) ( ) (iv) (v) 6( ) (vi) ( 5) (vii) ( 5) (viii) (i) y () 6( ). (i) 75 y (ii) 40 y 4 (iii) ( ) ( ) (iv) 4 4 (v) 7 ( ) ( 4) (vi) ( ) ( 5) (vii) ( 4) ( 4) ( 5) (viii) 4 (i) ( )( )( ) () 8( )( )( ) 6 Mthemtics Secondry Course

38 Specil Products nd Fctoriztion 4.7 MODULE -. (i), (ii), (iii), (v), (vii) nd (viii) 4.8. (i) (ii) 7 6 (iii) (iv) (v) (vi) 8 4 (vii) 5 (viii) 6. (i) 6 4 (ii) 8 4 (iii) (iv) (v) 4 (vi) ( ) 5 6 (vii) ( 9-7) - (viii). (i) (ii) 6 (iii) (iv) 0 (v) (vi) 5 (vii) 0 (viii) 8 (i) ± 5 () 4. (i) (ii) (iii) 6 (iv) (v) (vi) (vii) (viii) 6. (i) (ii) (iii) 7 (iv) 4 Mthemtics Secondry Course 7

39 MODULE - Specil Products nd Fctoriztion. (i) 5 (ii) (iii) (iv) 6 (v) 5 (vi) ANSWERS TO TERMINAL EXERCISE. (i) C (ii) A (iii) D (iv) A (v) D (vi) B (vii) C (viii) B (i) A () C. (i) m n (ii) y 4 4 (iii) 4 y 9y (iv) 9 0b 5b (v) 5 8y (vi) 8 5y 4 (vii) 0 (viii) 4z 4 4z 5 (i) () 0977 (i) b 0 (ii) 4 4z 5z (i) 7 y 6 ( 5 y 4 ) (ii) b( b) ( b) ( 9b ) (iii) ( b ) (iv) ( 4b ) (v) ( y y ) (vi) ( 4 9)( 4 9) (vii) ( 9)( 7) (viii) ( )( 9) (i) ( y)(7 6y) () ( ) (5 ) (i) ( y) ( y)( y 9y ) ( y 9y ) (ii) (5 4b )(5 4 0 b 6b 4 ) 6. (i) ( ) (ii) 0( ) 7. (i) ( y ) ( y y (ii) 4 y y 4 8. (i) (ii) (iii) 4 8 (iv) Mthemtics Secondry Course

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