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HCF and LCM

HCF and LCM 

1.          Define H.C.F.

Soln:  The HCF (Highest Common Factor) of two or more numbers/expressions is the greatest number/expression that exactly divides each of the given numbers/expressions.

2.          Define L.C.M.

Soln: The LCM (Lowest Common Multiple) of two or more given numbers/expressions is the smallest number/expression that is exactly divisible by each of the given numbers/expressions.

3.          If P and Q are two expressions such that P divides Q without remainder, then what will be their H.C.F?                   

Soln: If P divides Q without remainder, then their H.C.F will be P.

4.          Given any two expressions G and F if F divides G without remainder, what will be the L.C.M of G and F?               

Soln: If F divides G without remainder, then their L.C.M will be G.

5.          What is the H.C.F. of two expressions (x + 2) and (y + 3)?

Soln: In two expression (x + 2) and (y + 3), there is no common factor except 1. Hence F.C. F. of (x + 2) and (y + 3) is 1.

6.          What is the L.C.M of two expressions (x2 + a) and (x + a)?

Soln: The L.C.M of two expressions (x2 + a) and (x + a) is (x + a)(x2 + a).

7.    If H.C.F and the product of remaining factors of any two or more expressions are represented by P and Q respectively and R denotes the L.C.M of the expressions, then write the relation among P, Q and R.        

Soln: The relation among P, Q and R is R = P.Q.

8.   If G and L represent the HCF and LCM of two expressions P and Q respectively, write the relation among G, L, P and Q.

Soln:  The relation is as follows G × L = P (x) ´ Q (x). Ans.

9.       Find the LCM of :

                x(x – y) and y(x + y)              

Soln: Here, the first expression = x(x - y)

        Second expression = y(x + y)

         LCM = xy(x - y)(x + y) = xy(x2 - y2Ans.


1.          Find the HCF of :

         x4 + 4 and 3x2 – 6x + 6                                               

Soln: 1st exp. x4 + 4 = (x2)2 + 22

                           = (x2 + 2)2 – 4x2

                           = (x2 + 2)2 – (2x)2

                           = (x2 + 2 + 2x) (x2 + 2 – 2x)

                           = (x2 + 2x + 2) (x2 – 2x + 2)

         2nd exp.  3x2 – 6x  + 6 = 3 (x2 – 2x + 2)

         \ H.C.F. = x2 – 2x + 2.   Ans

2.          Find the HCF of:

         3x4 – 15x2 and 7x3 – 35x                                            

Soln:  Here, the first expression = 3x4 – 15x2

                                               = 3x2(x2 – 5)

                                              = 3 . x . x (x2 – 5)

        and, second expression  =             7x3 – 35x

                                             = 7x(x2 – 5)

          \     HCF = x (x2 – 5) Ans.

3.          Find the HCF of:

         x2 – 3x + 3y – y2 and y2 + xy – 3y                              

Soln: First expression = x2 – 3x + 3y – y2

                                  = x2 – y2 – 3x + 3y

                                  = (x – y) (x + y) – 3(x – y)

                                  = (x – y) (x + y – 3)

         Second expression = y2 + xy – 3y

                                     = y(y + x – 3)

                                     = y(x + y – 3)

         H.C.F. = x + y – 3 Ans.                                            

4.          Find the HCF of :

x3 – y3 and x2 + xy + y2                                               

Soln: Here, first expression = x3 - y3

                                        = (x - y)(x2 + xy + y2)

         Second expression = x2 + xy + y2

         H.C.F. = x2 + xy + y2 Ans.                                       

5.          Find the HCF of:

x3 + y3 and x2 – xy + y2                                               

Soln: Here, first expression = x3 + y3

                                        = (x + y)(x2 - xy + y2)

         Second expression = x2 - xy + y2

         H.C.F. = x2 - xy + y2 Ans.                                       

6.          Find the HCF of:

x2 – y2 and x2 – 2xy + y2                                                 

Soln: Here, first expression = x2 - y2

                                        = (x - y)(x + y)

         Second expression = x2 - 2xy + y2

                                    = (x - y)2

                                    = (x - y) (x - y)

         H.C.F. = x - y Ans.                                                  

7.          Find the H.C.F ofM

x4 – 64x and x2 + 4x + 16                                                

Soln: Here, first expression = x4 - 64x

                                        = x(x3 - 64)

                                        = x{x3 - (4)3}

                                        = x (x - 4){(x2 + x .4 + (4)2 }

                                        = x (x - 4)(x2 + 4x + 16)

         Second expression = x2 + 4x + 16

         H.C.F. = x2 + 4x + 16 Ans.                                      

8.          Find HCF of:

8m3 – 27n3; 8m2 + 12mn + 18n2                                                            

Soln: First expression= 8m3 – 27n3

                                = (2m)3 – (3n)3

                                = (2m – 3n) (4m2 + 6mn + 9n2)

         Second expression = 8m2 + 12mn + 18n2

                                   = 2(4m2 + 6mn+ 9n2)

         \   L.C.M. = 2(2m – 3n) (4m2 + 6mn + 9n2)

                            = 2 (8m3 – 27n3Ans.

9.          Find HCF of:

4x2 – 9y2; 8x3 + 27y3                                                        

Soln: Here, first expression = 4x2 - 9y2

                                        = (2x)2 - (3y)2

                                        = (2x - 3y) (2x + 3y)

         Second expression = 8x3 + 27y3

                                    = (2x)3 + (3y)3

                                    = (2x + 3y) {(2x)2 - 2x.3y + (3y)2}

                                    = (2x + 3y) (4x2 - 6xy + 9y2)

         H.C.F. = 2x + 3y Ans.                                              

10.       Find the H.C.F. of:

12x4 – 27x2y2 and 2x2 – xy  3y2                                    

Soln: Here, first expression = 12x4 - 27x2y2

                                        = 3x2(4x2 - 9y2)

                                        = 3x2{(2x)2 - (3y)2}

                                        = 3x2(2x - 3y) (2x + 3y)

         Second expression = 2x2 – xy –  3y2                                                                                                              = 2x2 – 3xy + 2xy  3y2                                                                  

                                       = x(2x - 3y) + y(2x + 3y)

                                       = (x - y)(2x - 3y)

         H.C.F. = 2x - 3y Ans.                                              

11.       Find HCF of

3m3 – 27m and m3 + 27                                                   

Soln: Here, first expression = 3m3 - 27m

                                        = 3m(m2 - 9)

                                        = 3m{(m)2 - (3)2}

                                        = 3x2(m - 3) (m + y)

         Second expression = m3 27                                                                                                                          = m3 + (3)3

                                       = (m + 3){m2 - m.3 + (3)2}

                                    =  (m + 3)(m2 - 3m + 9)

         H.C.F. = m + 3 Ans.                                                 

12.       Find the HCF of:

x2 – y2 + x + y and x2 – xy + x                        

Soln: Here, first expression = x2 – y2 + x + y                                                                                                              = (x - y) (x + y) + 1(x + y)

                                        = (x + y) (x - y + 1)

         Second expression = x2 – xy +  x                                                                                                                 = x(x - y + 1)

         H.C.F. = (x y + 1) Ans.                                         

13.       Find the HCF of:

x2 – y2 – 2x + 1 and x2 – xy – x                                       

Soln: Here, first expression = x2 – y2 - 2x + 1                                                                     

                                        = x2 - 2x + 1 – y2

                                        = (x - 1)2 - y2

                                        = (x - 1 + y) (x - 1 - y)

                                        = (x + y - 1) (x - 1)

         Second expression = x2 – xy -x                                                                                                                     = x(x - y - 1)

         H.C.F. = (x - 1) Ans.                                         

14.       Find the LCM of:

a4 + a2 + 1, a3 + a2 + a                                                 

Soln: First expression = a4 + a2 + 1

                                = (a2)2 + 12 + a2

                                = (a2 + 1)2 – 2a2.1 + a2

                                = (a2 + 1)2 – a2

                                = (a2 + 1 – a) (a2 + 1 + a)

                                = (a2 – a + 1) (a2 + a + 1)

         Second expression = a3 + a2 + a

                                    = a(a2 + a + 1)

         L.C.M. = a(a2 + a + 1) (a2 – a + 1) Ans

15.       Find the LCM of:

a3 + b3, a3b – a2b2 + ab3                                              

Soln: First expression= a3 + b3

                                = (a + b) (a2 - ab + b2)

         Second expression  = a3b – a2b2 + ab3

                                    = ab(a2 - ab + b2)

         L.C.M. = ab(a + b)(a2 – ab + b2) Ans

16.       Find the LCM of:

p4 + 4, 2p3 – 4p2 + 4p                                                  

Soln: First expression= p4 + 4

                                = (p2)2 + (2)2

                                = (p2+ 2)2 - 2.p2.2

                                = (p2 + 2)2 - 4p2

                                = (p2 + 2)2 - (2p)2

                                = (p2 + 2 - 2p) (p2 + 2 + 2p)

                                = (p2 - 2p + 2) (p2 + 2p + 2)

         Second expression   = 2p3 – 4p2 + 4p

                                    = 2p(p2 - 2p + 2)

         L.C.M. = 2p(p2 – 2p + 2) (p2 + 2p + 2) Ans

17.       Find the LCM of:

y3 – 1, y4 + y2 + 1                                                        

Soln: 1st exp.  y3 – 1 = y3 – 13 = (y – 1) (y2 + y + 1)

         2nd exp.  y4 + y2 + 1 = (y2)2 + (1)2 + y2

                                  = (y2 + 1)2 – 2y2 + y2

                                  = (y2 + 1)2 – y2

                                  = (y2 + 1 + y) (y2 + 1 – y)

                                  = (y2 + y + 1) (y2 – y + 1)

         \ L.C.M. = (y – 1) (y2 + y + 1) (y2 – y + 1) Ans.

18.       Find the LCM of :

x4 + x2 + 1 and x2 – x + 1                                            

Soln: 1st exp. = x4 + x2 + 1

                    = (x2)2 + (1)2 + x2

                    = (x2 + 1)2 – 2x2 + x2

                    = (x2 + 1)2 – x2

                    = (x2 + 1 + x) (x2 + 1 – x)

                    = (x2 + x + 1) (x2 – x + 1)

         2nd exp. = x2 – x + 1

         \ L.C.M. = (x2 + x + 1) (x2 – x + 1) = x4 + x2 + 1 Ans.

19.       Find the LCM of:

a3 + 8b3 and a2 – 2ab + 4b2

Soln: Here, first expression = a3 + 8b3

                                        = a3 + (2b)3

                                        = (a + 2b) {a2 - a.2b + (2b)2}

                                        = (a + 2b) (a2 - 2ab + 4b2)

         Second expression = a2 – 2ab + 4b2

         L.C.M. = (a + 2b) (a2 - 2ab + 4b2) = a3 + 8b3 Ans.  

20.       Find the LCM of:

(x + y)2 – 4xy and x2 + xy – 2y2   

Soln: 1st exp. (x + y)2 – 4xy = x2 + 2xy + y2 – 4xy

                                     = x2 – 2xy + y2

                                     = (x – y)2

         2nd exp.  x2 + xy – 2y2 = x2 + 2xy – xy – 2y2

                                     = x (x + 2y) – y (x + 2y)

                                     = (x + 2y) (x – y)

         \ L.C.M. = (x – y) (x – y) (x + 2y) = (x – y)2 (x + 2y)  Ans.                                                                                           

30.       Find the HCF of:

x2 + y2 + 2xy – 1, y2 – x2 + 2y + 1 and x2 – y2 + 2x + 1                                                                                                   

Soln: 1st exp.  = x2 + y2 + 2xy – 1

= (x + y)– 12

= (x + y + 1) (x + y – 1)

         2nd  exp.  = y2 – x2 + 2y + 1

= (y2 + 2y + 1) – x2

= (y + 1)2 – x2

= (y + 1 + x) (y + 1 – x)

= (x + y + 1) (–x + y + 1)

         3rd exp.  = x2 – y2 + 2x + 1

                      = (x2 + 2x + 1) – y2

                      = (x +1)2 – y2 

                      = (x + 1 + y) (x + 1 – y)

                      = (x + y + 1) (x – y + 1)

         \ H.C.F. = (x + y + 1)

31.       Find the HCF of:

5x2 – 125, x2 – 10x + 25 and 2x2 – 10x                      

Soln: 1st exp.  = 5x2 – 125

= 5(x2 - 25)

= 5{x2 - (5)2}

= 5(x - 5)(x + 5)

         2nd  exp.  = x2 – 10x + 25

= x2 – 5x - 5x + 25

= x(x - 5) – 5(x - 5)

= (x- 5 (x + 5)

         3rd exp.  = 2x2 – 10x

                      = 2x(x- 5)

         \ H.C.F. = (x - 5) Ans

32.       Find the HCF of:

1 + 4x + 4x2 – 16x4 and 1 + 2x – 8x3 – 16x4                                                                            

Soln: 1st expression= 1 + 4x + 4x2 – 16x4

                            = (1 + 2x)2 – 16x4

                            = (1 + 2x)2 – (4x2)2

                            = {(1 + 2x) + 4x2} {(1 + 2x) – 4x2}

                            = (1 + 2x + 4x2) (1 + 2x – 4x2)

         2nd expression = 1 + 2x – 8x3 – 16x4

                            = 1 – 8x3 + 2x – 16x4

                            = (1 – 8x3) + 2x(1 – 8x3)

                            = (1 – 8x3) (1 + 2x)

                            = {1 – (2x)3} (1 + 2x)

                            = (1 – 2x) (1 + 2x + 4x2) (1 + 2x)

         H.C.F. = 1 + 2x + 4x2 Ans.

33.       Find the HCF of:

y– 1, y4 + y2 + 1 and y3 + 1 + 2y2 + 2y

Soln:  1st expression =  y3 – 1

= (y – 1) (y2 + y + 1)

          2nd expression = y4 + y2 + 1

       = (y2+ 1)2 – 2y2 + y2 

       = (y2 + 1)2 – y2

       = (y2 + 1 + y) (y2 + 1 – y)

       = (y2 + y + 1) (y2 – y + 1)

          3rd expression: = y3 + 1 + 2y2 + 2y

       = (y + 1) (y2 – y + 1) + 2y (y + 1)

       = (y + 1) (y2 – y + 1 + 2y)

       = (y + 1) (y2 + y + 1)

          \ H.C.F. = (y2 + y + 1) Ans.

34.       Find the LCM of:

x3 + 7x2 + 12x, x3 + 64 and 3x2 + 27x + 60

Soln: 1st expression = x3 + 7x2 + 12x

                            = x(x2 + 7x + 12)

                            = x{x2 + (4 + 3) x + 12}

                            = x{x2 + 4x + 3x + 12}

                            = x{x(x + 4) + 3(x + 4)}

                            = x (x + 4) (x + 3)

         2nd expression= x3 + 64

                            = x3 + (4)3

                            = (x + 4) (x2 – 4x + 16)

         3rd expression= 3x2 + 27x + 60

                            = 3(x2 + 9x + 20)

                            = 3{x2 + 5x + 4x + 20}

                            = 3{x(x + 5) + 4(x + 5)}

                            = 3(x +5) (x + 4)

         L.C.M. = 3x(x + 3) (x + 4) (x + 5) (x2 – 4x + 16)  Ans

35.       Find the LCM of:

8x3 + 27y3, 8x3 – 27y3 and 16x4 + 36x2y2 + 81y4       

             1st expression  = 8x3 + 27y3

                                                          = (2x)3 + (3y)3

                                      = (2x + 3y) {(2x)2 – 2x.3y + (3y)2}

                                      = (2x + 3y) (4x2 – 6xy + 9y2)

             2nd expression = 8x3 - 27y3

                                                          = (2x)3 + (3y)3

                                      = (2x - 3y) {(2x)2 + 2x.3y + (3y)2}

                                      = (2x - 3y) (4x2 + 6xy + 9y2)

             3rd expression  = 16x4 + 36x2y2 + 81y4

                                      = (4x2 + 9y2)2 – (6xy)2

                                      = (4x2 + 9y2 + 6xy) (4x2 + 9y2 – 6xy)

                                      = (4x2 + 6xy + 9y2) (4x2 – 6xy + 9y2)

             \ L.C.M. = (4x2 + 6xy + 9y2) (4x2 – 6xy + 9y2) (2x - 3y) (2x + 3y)

                            =  (8x3 + 27y3) (8x3 – 27y3) Ans

36.       Find the LCM of:

         x3 – 9x, x4 – 2x3 – 3x2 and x3 – 27                                         

Soln: 1st expression = x3 – 9x

                            = x(x2 – 9)

                            = x(x + 3) (x – 3)

         2nd expression = x4 – 2x3 – 3x2

                            = x2(x2 – 2x – 3)

                            = x2{x2 – 3x + x – 3}

                            = x2{x(x – 3) + 1(x – 3)}

                            = x2(x – 3) (x + 1)

         3rd expression = x3 – 27

                            = x3 – (3)3

                            = (x – 3) (x2 + 3x + 9)

         \   L.C.M.  = x.x (x + 1) (x + 3) (x – 3) (x2 + 3x + 9)

                            = x2(x + 1) (x2 – 9) (x2 + 3x + 9) Ans.

37.       Find the LCM of:

x4 + x2 + 1, x4 – x and 2x3 + 2x2 + 2x

Soln:  1st expression = x4 + x2 + 1

       = (x2+ 1)2 – 2x2 + x2 

       = (x2 + 1)2 – x2

       = (x2 + 1 + x) (x2 + 1 – x)

                            = (x2 + x + 1) (x2 – x + 1)

          2nd expression =  x4 – x

                                = x(x3 - 1)

= x(x – 1) (x2 + x + 1)

                                  3rd expression =  2x3 + 2x2 + 2x

  =2x(x2 + x + 1)

          \ L.C.M. = 2x(x – 1) (x2 + x + 1) (x2 – x + 1) Ans.

38.       Find the LCM of:

a2 – b2 + 2ab – c2, (a + b)2 – c2, (a + b – c)2

Soln:  1st expression = a2 – b2 + 2ab – c2                                                                      

          2nd expression                             = (a + b)2 – c2

                                  = (a + b + c) (a + b – c)

               3rd expression = (a + b – c)2

                                  = (a + b – c) (a + b – c)

          \            LCM = (a + b + c) (a + b – c) (a + b – c) (a2 – b2 + 2ab – c2)

                                  = (a + b + c) (a + b – c)2 (a– b + 2ab – c2Ans.





30.       Find the LCM of:

ab – ac + bc – b2, bc – ab + ac – c2 and ac – bc + ab – a2

Soln:  1st expression = ab – ac + bc – b2

                                  = ab – ac - b2 + bc

                                  = a(b - c) - b(b - c)

                                  = (b - c) (a - b)

          2nd expression    = bc – ab + ac – c2

                                  = bc – ab - c2 + ac

                                  = b(c - a) - c(c - a)

                                  = (c - a) (b -c)

                                                3rd expression = ac – bc + ab – a2

                                                        = ac – bc - a2 + ab

                                                      = c(a - b) - a(a - b)

                                                = (a - b) (c - a)

          \ LCM   = (a – b) (b – c) (c – a) Ans.

31.       Find the H.C.F of:

         (p + r) (p – r) + q(2p + q), (q + p) (q – p) + r(2q + r) and
(r + q) (r – q) + p(2r + p)                                            

Soln: Here, 1st expression = (p + r) (p – r) + q(2p + q)

                                      = p2 - r2 + 2pq + q2

                                      = (p + q)2 - r2

                                      = (p + q - r) (p + q + r)

                  2nd expression = (q + p) (q – p) + r(2q + r)

                                      = q2 - p2 + 2qr + r2

                                      = (q + r)2 - p2

                                      = (q + r - p) (q + r + p)

                  3rd expression = (r + q) (r – q) + p(2p + p)

                                      = r2 - q2 + 2pq + p2

                                      = (q + r)2 - p2

                                      = (q + r - p) (q + r + p)

         \ H.C.F. = (p + q + r) Ans.

32.       Find the L.C.M of:

         a(a + c) – b(b + c), b(a + b) – c(c + a) and c(b + c) – a(a + b)

Soln: First exp. = a(a + c) – b(b + c)

                        = a2 + ac - b2 - bc

                        = a2 - b2 + ac - bc

                        = (a + b) (a - b) + c(a - b)

                        = (a - b) (a + b + c)

         Second exp. = b(a + b) – c(c + a)

                            = ab + b2 - c2 - ac

                            = b2 - c2 + ab - ac

                            = (b + c) (b - c) + a(b - c)

                            = (b - c) (b + c + a)

         Third exp.    = c(b + c) – a(a + b)

                            = bc + c2 - a2 - ab

                            = c2 - a2 + bc - ab

                            = (c + a) (c - a) + b(c - a)

                            = (c - a) (c + a + b)

         \ L.C.M. = (a + b + c) (a – b)(b – c)(c – a) = (a – b)(b – c)(c – a)(a + b + c)

                                                                                    

33.       L3 + M3 = P3 + Q3.If H is the HCF and L, the LCM of two quantities P and Q, and if L + M = P + Q, then prove that: L3 + M3 = P3 + Q3

Soln: Here,  H.C.F. = H, L.C.M. = L, L + M = P + Q

         We know that, product of H.C.F. and L.C.M. = product of two expressions

         or,     HL = PQ

         Now, we have

                   L + M = P + Q

         Cubing on both sides,

                   (L + M)3 = (P + Q)3

         or,     L3 + M3 + 3LM (L + M) = P3 + Q3 + 3PQ (P + Q)

         or,     L3 + M3 + 3PQ (P + Q) = P+ Q3 + 3PQ (P + Q)

         \      L3 + M3 = P3 + Q3.  Proved.

34.       Find the L.C.M. of):

1 – x2, 1 – 6x + 5x2, 1 – 4x – 20x3 – 25x4                                                       

Soln: 1st exp.  = 1 – x2

                      = (1 + x) (1 – x)

         2nd exp. = 1 – 6x + 5x2

                               = 1 – 5x – x + 5x2

                    = (1 – 5x) – x (1 – 5x)

                    = (1 – 5x) (1 – x)

         3rd exp. = 1 – 4x – 20x3 – 25x4

                                = (1 – 25x4) – 4x – 20x3

                    = (1 + 5x2) (1 – 5x2) – 4x (1 + 5x2)

                    = (1 + 5x2) (1 – 5x2 – 4x)

                    = (1 + 5x2) (1 – 4x – 5x2)

                    = (1 + 5x2) (1 – 5x + x – 5x2)

                    = (1 + 5x2) {(1 – 5x) + x (1 – 5x)}

                    = (1 +5x2) (1 – 5x) (1 + x)

         \ L.C.M. = (1 + x) (1 – x) (1 – 5x) (1 + 5x2)

                      = (1 – x2) (1 – 5x) (1 + 5x2)   Ans.

35.       Find the LCM of:

m2 + m + 1, m3 – 1, m6 – 1                                          

Soln: 1st exp. = m2 + m + 1

         2nd exp.  = m3 – 1

= m3 – 13

= (m – 1) (m2 + m + 1)

         3rd exp. = m6 – 1

= (m3)2 – 12

= (m3 + 1) (m3 – 1)

= (m3 + 1) (m – 1) (m2 + m + 1)

         \ L.C.M. = (m – 1) (m2 + m + 1) (m3 + 1) = (m3 – 1) (m3 + 1) = (m6 – 1)   Ans.

36.       Find the HCF of) :

         a4 – 3a3 + 3a – 1, a6 – 1 and a3 – a                             

Soln: Here,

         1st expn.= a4 – 3a3 + 3a – 1

                     = a4 – a3 – 2a3 + 2a2 – 2a2 + 2a + a – 1

                     = a3(a – 1) – 2a2(a – 1) – 2a(a – 1) + 1(a – 1)

                     = (a – 1) (a3 – 2a2 – 2a + 1)

                     = (a – 1) {a3 + a2 – 3a2 – 3a + a + 1}

                     = (a – 1) {a2(a + 1) – 3a(a + 1) + 1(a + 1)}

                     = (a – 1) (a + 1) (a2 – 3a + 1)

                     = (a – 1) (a + 1) (a2 – 3a+ 1)

         2nd expn. = a6 – 1

                     = (a3)2 – (1)2                                                     

                     = (a3 – 1) (a3 + 1)

                     = (a – 1) (a2 + a + 1) (a + 1) (a2 – a + 1)

         3rd expn. = a3 – a

                    = a(a2 – 1)

                     = a(a – 1) (a + 1)

         \        H.C.F. = (a – 1) (a + 1) = a2 – 1 Ans.

37.        (Find the LCM of):                                                   

         9x4 – 28x2 + 3, 27x4 – 12x2 + 1 and x4 – 6x2 + 9

Soln: 1st expression = 9x4 – 28x2 + 3

                                = 9x4 – 27x2 – x2 + 3

                                = 9x2 (x2 – 3) – 1(x2 – 3)

                               = (x2 – 3) (9x2 – 1)

                                = (x2 – 3) {(3x)2 – (1)2}

                                = (x2 – 3) (3x – 1) (3x + 1)

         2nd expression                                   = 27x4 – 12x2 + 1

                               = 27x4 – 9x2 – 3x2 + 1

                                = 9x2(3x2 – 1) – 1(3x2 – 1)

                                = (3x2 – 1) (9x2 – 1)

                                = (3x2 – 1) {(3x)2 – (1)2}

                                = (3x2 – 1) (3x – 1) (3x + 1)

         3rd expression      = x4 – 6x2 + 9

                                = (x2)2 – 2.x2.3 + (3)2

                                = (x2 – 3)2

              LCM  = (x2 – 3) (3x – 1) (3x + 1) (3x2 – 1) (x2 – 3)

                   = (9x2 – 1) (3x2 – 1) (x2 – 3)2  Ans.

38.        (Find the HCF of):

         x3 – 64y3, x2 – 6xy + 8y2 and x2 – 16y2             

Soln: First expression   = x3 – 64y= x3 – (4y)3

                                      = (x – 4y) (x2 + x × 4y + 16y2)

                                      = (x – 4y) (x2 + 4xy + 16y2)

         Second expression                     = x2 – 6xy + 8y2

                                      = x2 – 4xy – 2xy + 8y2

                                      = x (x – 4y) – 2y (x – 4y)

                                      = (x – 2y) (x – 4y)

         Third expression  = x2 – 16y= x2 – (4y)

                                                         = (x + 4y) (x – 4y)

                         H.C.F. = (x – 4y) Ans.

39.        (Find the LCM of):

         x2 – 7xy + 12y2, x2 + 2xy – 15y2 and x2 + xy – 20y2

Soln: 1st expression                                    = x2 – 7xy + 12y2

                                = x2 – 4xy – 3xy + 12y2

                                = x(x – 4y) – 3y(x – 4y)

                                = (x – 4y) (x – 3y)

         2nd expression                                   = x2 + 2xy – 15y2

                                = x2 + 5xy – 3xy – 15y2

                                                = x(x + 5y) – 3y(x + 5y)

                                = (x + 5y) (x – 3y)

         3rd expression                                     =  x2 + xy – 20y2

                                = x2 + 5xy – 4xy – 20y2

                                = x(x + 5y) – 4y(x + 5y)

                                = (x + 5y) (x – 4y)

         \ L.C.M. = (x – 3y) (x – 4y) (x + 5y)   Ans.

40.       Find the H.C.F. of) :

         b2 (b2 + 4bc + 4c2), b5 + 8b2c3 and 3b4 + b3c – 10b2c2

Soln: Here,

         1st exp. =b(b2 + 4bc + 4c2)

                      = b2 (b + 2c)2

              2nd exp. = b5 + 8b2c3

                           = b2 (b3 + 8c3)

                      = b2 {b3 + (2c)3}

                      = b2 (b + 2c) (b2 – 2bc + 4c2)

         3rd exp.  = 3b4 + b3c – 10b2c2

                      = b2 (3b2 + bc – 10c2)

                      = b2 {3b2 + 6bc – 5bc – 10c2}

                      = b2 {3b (b + 2c) – 5c(b + 2c)}

                      = b2 (b + 2c) (3b – 5c)

         \         HCF = b2 (b + 2c) Ans.

41.       Find the H.C.F. of) :

         p2 + 4pq + 4q2, p4 + 8pq3 and 3p4 – 10p2q2 + p3q       

Soln: 1st exp.=                                                 p2 + 4pq + 4q2

= (p + 2q)2

= (p + 2q) (p + 2q)

         2nd exp.=p4 + 8pq3

                    = p(p3 + 8q3)

                    = p{p3 + (2q)3}

                    = p(p + 2q) (p2 – 2pq + 4q2)

         3rd exp. = 3p4 – 10p2q2 + p3q

                      = (3p4 + p3q – 10p2q2)

                      = p2(3p2 + pq – 10q2)

                      = p2(3p2 + 6pq – 5pq – 10q2)

                      = p{3p(p + 2q) – 5q(p + 2q)}

                      = p2(p + 2q) (3p – 5q)

         \      HCF = (p + 2q) Ans.

42.        (Find the H.C.F. of) :

         v2(v2 + 4vw + 4w2), v5 + 8v2w3 and 3v4 + v3w – 10v2w2

Soln: Here,

       First expression   = v2(v2 + 4vw + 4w2)

                                      = v2(v2 + 2vw + 2vw + 4w2)

                                      = v2{v(v + 2w) + 2w(v + 2w)}

                                      = v2(v + 2w) (v + 2w)

         Second expression  = v5 + 8v2w3

                                                         = v2(v3 + 8w3) = v2{v3 + (2w)3}

                                      = v2(v + 2w) (v2 – 2vw + 4w2)

         Third expression = 3v4 + v4w – 10v2w2

                                      = v2(3v2 + vw – 10w2)

                                      = v2(3v2 + 6vw – 5vw – 10w2)

                                      = v2{3v(v + 2w) – 5w(v + 2w)}

                                      = v2(v + 2w) (3v – 5w)

         Hence, HCF = v2 (v + 2w) Ans.

43.        (Find the H.C.F. of):

         a2 + 4ab + 4b2, a4 + 8ab3 and 3a4 – 10a2b2 + a3b       

Soln: Here,

       First expression                          = a2 + 4ab + 4b2

                                   = a2 + 2ab + 2ab + 4b2

                                   = a(a + 2b) + 2b(a + 2b)

                                   = (a + 2b) (a + 2b)

         Second expression = a4 + 8ab3

                                   = a(a3 + 8b3)

                                   = a{a3 + (2b)3}

                                   = a(a + 2b) {a2 – a.2b + (2b)2}

                                   = a(a + 2b) (a2 – 2ab + 4b2)

         Third expression                    = 3a4 – 10a2b2 + a3b

                                   = a2(3a2 + ab – 10b2)

                                   = a2(3a2 + 6ab – 5ab – 10b2)

                                   = a2 {3a(a + 2b) – 5b(a + 2b)}

                                   = a2(a + 2b) (3a – 5b)

         Hence, HCF = (a + 2b) Ans.

44.         (Find the H.C.F. of) :

         x3 + 4x2y + 4xy2, x4 + 8xy3 and 3x4 + x3y – 10x2y2

Soln: Here,

       First expression   = x+ 4x2y + 4xy2

                                      = x(x2 + 4xy + 4y2)

                                      = x (x2 + 2xy + 2xy + 4y2)

                                      = x{x(x + 2y) + 2y(x + 2y)}

                                      = x(x + 2y) (x + 2y)

         Second expression                             = x4 + 8xy3

                                                         = x(x3 + 8y3)                                   

                                      = x{x3 + (2y)3}

                                      = x(x + 2y) {x2 – x.2y + (2y2})

                                      = x(x + 2y) (x2 – 2xy + 4y2)

         Third expression = 3x4 + x3y – 10x2y2

                                      = x2(3x2 + xy – 10y2)

                                      = x2(3x2 + 6xy – 5xy – 10y2)

                                      = x2{3x(x + 2y) – 5y(x + 2y)}

                                      = x2(x + 2y) (3x – 5y)

         Hence, H.C.F. = x(x + 2y) Ans.

 


Questions For practice(HCF)





Problem for Practice 














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