HCF AND LCM – EXERCISE 3.3.4 – Class 9

  1. Find the LCM of the following:

(i) 2x + 6, x2 + 3x

Solution:

2x + 6 = 2 (x + 3)

x2 + 3x = x (x + 3)

LCM = 2x (x + 3)

 

(ii) x2y + xy2, x 2+ xy

Solution:

x2y + xy2 = xy (x + y)

x 2+ xy = x ( x+ y)

LCM = xy (x + y)

 

(iii) 3x2 – 75, 2x3 + 250

Solution:

3x2 – 75 = 3 (x2 – 25)

= 3 (x2 – 52)

= 3 (x + 5) (x – 5)

2x3 + 250 = 2 (x3 + 125)

LCM = 2 x 3 (x + 5) (x3 + 125)

= 6 (x + 5) (x3 + 125)

 

(iv) a2 – 1, a4 – 1, a8 – 1

Solution:

a2 – 1 = (a + 1) (a – 1)

a4 – 1 = (a2 + 1) (a2 – 1)

= (a2 + 1) (a + 1) (a – 1)

a8 – 1 = (a4 + 1) (a4 – 1)

= (a4 + 1) [(a2)2 – 12]

= (a4 + 1) (a2 + 1) (a2 – 1)

= (a4 + 1) (a2 + 1) (a + 1) (a – 1)

LCM = (a + 1) (a – 1) (a2 + 1) (a4 + 1)

 

(v) m2 – n2 , 3m2 – 3mn

Solution:

m2 – n2 = (m + n) (m – n)

3m2 – 3mn = 3m (m – n)

LCM = 3m (m + n) (m – n)

 

(vi) 5(y2 – z2), y2 + 2yz + z2

Solution:

5(y2 – z2) = 5 (y – z) (y + z)

y2 + 2yz + z2 = (y + z)2 [(a + b)2 = a2 + 2ab + b2]

LCM = 5 (y – z) (y + z)2

 

(vii) x3 + 8, x2 – 4

Solution:

x3 + 8 = x3 + 23

= (x + 2) (x2 – 2x +4)

x2 – 4 = x2 + 22

= (x + 2) (x – 2)

LCM = (x + 2) (x – 2) (x2 – 2x +4) = (x+2)(x3+8)

 

(viii) 3(a + b)2, 5(a – b )2, 2(a2 – b2)

Solution:

3(a + b)2 = 3(a + b) (a + b)

5(a – b)2 = 5(a – b ) (a – b)

2(a2 – b2) = 2(a + b) (a – b)

LCM = 2 x 3 x 5 (a + b)2 (a – b)2

LCM =30 (a + b)2 (a – b)2 = 30(a2-b2)2

 

(ix) 8x3 – y3, ab (4x2 + 2xy + y2), bc (4x2 – y2)

Solution:

8x3 – y3 = (2x)3 – y3

= (2x – y) [(2x)2 + 2.x.y + y2]

= (2x – y) (4x2 + 2xy + y2)

ab (4x2 + 2xy + y2)

bc (4x2 – y2) = bc [(2x)2 – y2] = bc (2x + y) (2x – y)

LCM = abc (2x – y) (2x + y) (4x2 + 2xy + y2) =  abc(2x+y)(8x3-y3)

 

(x) 21(x – 1)2 , 35 (x4 – x2) , 14 (x4 – x)

Solution:

21(x – 1)2 = 3 x 7 (x – 1)2

35 (x4 – x2) = 5 x 7x2 (x2 – 1)

= 5 x 7x2 (x + 1) (x – 1)

14 (x4 – x) = 2 x 7x (x3 – 1)

= 2 x 7x (x3 – 1)

= 2×7x (x-1)(x2 + x +1)

LCM = 2 x 3 x 5 x 7×2 (x + 1)(x-1)2 (x2 + x +1)

= 210x2(x+1)(x-1)(x3-1) = 210x2(x2-1)(x3-1)


  1. Find the LCM of the following:

(i) x2 – 3x – 4, x2 +2x – 24

Solution:

x2 – 3x – 4

= x2 – 4x + x – 4

= x (x – 4) +1 (x – 4)

= (x – 4) (x + 1)

x2 +2x – 24

= x2 +6x – 4x – 24

= x (x + 6) – 4 (x + 6)

= (x + 6) (x – 4)

LCM = (x – 4) (x + 1) (x + 6)

 

(ii) x2 + 4x + 4, x2 + 5x + 6

x2 + 4x + 4

= x2 + 2x + 2x + 4

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

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

x2 + 5x + 6

= x2 + 2x + 3x + 6

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

= (x + 2) (x +3)

LCM = (x + 2)2 (x +3)

 

(iii) – x2 – x + 6, – x2 + x + 2

– x2 – x + 6

= – (x2 + x – 6)

= (x2 + 3x – 2x – 6)

=[x(x + 3) – 2(x + 3)]

= (x – 2) (x + 3)

– x2 + x + 2

= – (x2 – x – 2)

= – x2+ 2x – x + 2]

= x(-x + 2) + 1(-x + 2)]

= (-x + 2) (x + 1) = (x+1)(2 – x)

LCM = (2 – x) (x + 1) (x + 3)

 

(iv) 6m² – 3m – 45, 6m² + 11m – 10

Solution:

6m2 – 3m – 45

= 3(2m2 – 3m – 15)

= 3(2m2 – 6m + 5m – 15)

= 3[(2m (m – 3) + 5 (m – 3)]

= 3(m – 3)(2m + 5)

6m2 + 11m – 10

= 6m2 + 15m – 4m – 10

= 3m (2m + 5) – 2 (2m + 5)

= (2m + 5) (3m – 2)

LCM = 3(m – 3) (2m + 5) (3m – 2)

 

(v) 10x3 + 6x2 – 28x , 9x3 + 15x2 – 6x

Solution:

10x3 + 6x2 – 28x

= 2x (5x2 + 3x – 14)

= 2x (5x2 + 10x – 7x – 14)

= 2x [5x (x + 2) – 7 (x + 2)]

= 2x (x + 2) (5x – 7)

9x3 + 15x2 – 6x

= 3x (3x2 + 5x – 2)

= 3x (3x2 + 6x – x – 2)

= 3x [3x (x + 2) – 1 (x + 2)]

= 3x (x + 2) (3x – 1)

LCM = 2 x 3 x x (x + 2) (3x – 1) (5x – 7)

= 6x (x + 2) (3x – 1) (5x – 7)

 

(vi) 6a3 + 60a2 + 150a, 3a4 + 12a3 – 15a2

6a3 + 60a2 + 150a

= 6a (a3 + 10a2 + 25)

= 6a (a3 + 2.5a2 + 52)

= 6a (a + 5)2

3a4 + 12a3 – 15a2

= 3a2 (a2 + 4a – 5)

= 3a2 (a2 + 5a – a – 5)

= 3a2 [a (a + 5) – 1 (a + 5)]

= 3a2 (a + 5) (a – 1)

LCM = 6a2 (a + 5)2 (a – 1)

 

(vii) 12x4 + 324x, 36x3 + 90x2 – 54x

Solution:

12x4 + 324x

= 12x (x3 + 27)

= 12x (x3 + 33)

= 12x (x + 3) (x2 – 3x + 9)

= 22 x 3x (x + 3) (x2 – 3x + 9)

36x3 + 90x2 – 54x

= 18x (2x2 + 5x – 3)

= 18x (2x2 + 6x – x – 3)

= 18x [2x (x + 3) – 1 (x + 3)

= 18x (x + 3) (2x – 1)

= 2 x 33 x (x + 3) (2x – 1)

LCM = 22 x 33 x (x + 3) (2x – 1) (x2 – 3x + 9)

= 36x (x3 – 27) (2x – 1)

 

(viii) a2 – 3a + 2, a3 – a2 – 4a + 4, a (a3 – 8)

Solution:

a2 – 3a + 2 = a2 – 2a – a + 2

= a (a – 2) – 1 (a – 2)

= (a – 1) (a – 2)

a3 – a2 – 4a + 4 = a2 (a – 1) – 4 (a – 1)

= (a – 1) (a2 – 4)

= (a – 1) (a2 – 22)

= (a – 1) (a – 2) (a + 2)

a (a3 – 8) = a (a3 – 23)

= a (a – 2) (a2 + 2a + 4)

LCM = a (a – 1) (a – 2) (a + 2) (a2 + 2a + 4)

= a (a – 1) (a + 2) (a3 – 8)

 

(ix) 4x3 + 4x2 – x – 1, 8x3 – 1, 8x2 – 2x – 1

4x3 + 4x2 – x – 1

= 4x2(x + 1) – 1 (x + 1)

= (x +1) (4x2 – 1)

= (x – 1) [(2x)2 – 1]

= (x – 1) (2x + 1) (2x – 1)

8x3 – 1 = (2x)3 – 1

= (2x – 1) [(2x)2 + 2x.1 + 12]

=(2x – 1) (4x2 + 2x + 1)

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

= (x + 1) (2x + 1) (8x3 – 1)

 

(x) m2 – 9m – 22, m2 – 8m – 33, m2 + 5m + 6

Solution:

m2 – 9m – 22

= m2 – 11m + 2m – 22

= m (m – 11) + 2 (m – 11)

= (m – 11) (m + 2)

m2 – 8m – 33

= m2 – 11m + 3m – 33

= m (m – 11) + 3 (m – 11)

= (m – 11) (m + 3)

m2 + 5m + 6 = m2 + 3m + 2m + 6

= m (m + 3) + 2 (m + 3)

= (m + 3) (m +2)

LCM = (m – 11) (m + 2) (m + 3)

 

(xi) 6 (x2 + 2xy – 3y3), 4(x2 – 3xy + 2y2), 8(x2 + xy – 6y2)

Solution:

6 (x2 + 2xy – 3y3) = 6 (x2 + 3xy – xy – 3y2)

= 6 [x (x + 3y) – y (x + 3y)]

= 6 (x + 3y) (x – y)

= 2 x 3 (x + 3y) (x – y)

= 4 (x2 – 3xy +2 y2)

4 (x2 – 3xy + 2y2) = 4 (x2 – 2xy – xy + 2y2)

= 4 [x (x – 2y) –y (x – 2y)]

= 4(x – 2y) (x – y)

= 22 (x – 2y) (x – y)

8(x2 + xy – 6y2) = 8(x2 + 3xy – 2xy – 6y2)

= 8 [x (x + 3y) – 2y (x +3y)]

= 8 (x + 3y) (x – 2y)

= 23 (x + 3y) (x – 2y)

LCM = 23 x 3 (x + 3y) (x – y) (x – 2y)

= 24 (x – y) (x – 2y) (x + 3y)

 

(xii) pq2 (x2 + x – 20), p2q(x2 – 3x – 4), p2q2 (x2 + 2x + 1)

pq2 (x2 + x – 20)

= pq2 (x2 + 5x – 4x – 20)

= pq2 [x (x + 5) – 4(x + 5)]

= pq2 (x + 5) (x – 4)

p2q(x2 – 3x – 4)

= p2q(x2 – 4x + x – 4)

= pq2 [x (x – 4) + 1 (x – 4)]

= pq2 (x – 4) (x + 1)

p2q2 (x2 + 2x + 1) = p2q2 (x2 + 2x + 1)

= p2q2 (x + 1)2

= p2q2 (x + 1)2 [(a+ b)2 = a2 + 2ab + b2]

LCM = p2q2 (x + 5) (x – 4) (x + 1)2

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