Find HCF LCM of the following monomials:

1) \ \ ab^2 \ \ and\ \ a^2 b

HCF = ab

LCM = a^2 b^2

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2) \ \ a^3 b^2 \ \ and\ \ a^2 b^4

HCM = a^2 b^2

LCM = a^2 b^4

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3) \ \ 4x^2 y^3 \ \ and\ \ 6xy^4

HCF = 2xy^3

LCM = 12x^2 y^4

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4) \ \ 6abc \ \ and\ \ 9bc^2 d

HCM = 3bc

LCM = 18abc^2 d

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5) \ \ 2m^2 n^3, \ \ 3mn^2 \ \ and\ \ 4m^3 n

HCM = mn

LCM = 12m^3 n^3

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6) \ \ 5x^3 y^2, 10x^2 z^2 \ \ and\ \ 15y^3 z^3

HCF = 5x^2

LCM = 30x^3 y^3 z^3

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7) \ \ 6x^2 y^2 z^4, 9x^4 y^5 z \ \ and\ \ 12xy^2 z^3

HCF = 3xy^2 z

LCM= 36x^4 y^5 z^4

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Find HCF and LCM of following polynomials

1) \ \ x^2-a^2 \ \ and\ \ x^2-ax

We have

x^2-a^2=(x-a)(x+a)=x^2-ax x(x-a)

Therefore

HCF = (x-a)

LCM= x (x-a)(x+a)

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2) \ \ 9a^2-16b^2 \ \ and\ \ 6a^2+8ab

We have

9a^2-16b^2=(3a-4b)(3a+4b)

6a^2+8ab=2a(3a+4b)

Therefore

HCF =(3a+4b)

LCM = 2a(3a+4b)(3a-4b)

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3) \ \ x^3-16x \ \ and\ \ x^3+2x^2-24x

We have

x^3-16x=x(x-4)(x+4)

x^3+2x^2-24x=x (x^2+2x-24)=x(x+6)(x-4)

HCF = x(x-4)

LCM = x(x-4)(x+6)(x+4)

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4) \ \ x^2+x-12 and x^2-6x+9

We have

x^2+x-12=(x+4)(x-3)

x^2-6x+9=(x-3)(x-3)

HCF= (x-3)

LCM = (x-3)(x-3)(x+4)

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5) \ \ x^2-36 \ \ and\ \ 2x^2-15x+18

We have

x^2-36=(x-6)(x+6)

2x^2-15x+18= 2x^2-12x-3x+18= 2x(x-6)-3(x-6)= (2x-3)(x-6)

Therefore

HCF = (x-6)

LCM = (x-6)(2x-3)(x+6)

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6) \ \ 6a^2 b (a^2-b^2 ) \ \ and\ \ 4ab^2 (a+b)^2

We have

6a^2 b (a^2-b^2 )= 6a^2 b (a-b)(a+b)

4ab^2 (a+b)^2=4a^2 b^2 (a+b)(a-b)

Therefore

HCF=2ab(a+b)

LCM=12a^2 b^2 (a+b)(a-b)

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7)\ \ xy+y^2, xy-y^2 \ \ and\ \ x^2 y^2-y^4

We have

xy+y^2=y(x+y)

xy-y^2=y(x-y)

x^2 y^2-y^4=y^2 (x^2-y^2 )= y^2 (x-y)(x+y)

HCF = y

LCM = y^2 (x-y)(x+y)

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8)\ \ x^2+3x, x^2+5x+6 \ \ and\ \ x^2+4x+3

We have

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

x^2+5x+6=(x+12)(x+3)

x^2+4x+3=(x+1)(x+3)

Therefore

HCF = (x+3)

LCM= x(x+1)(x+2)(x+3)

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9)\ \ x^2+3x-4, x^3-2x^2+x \ \ and\ \ x^2+2x-3

We have

x^2+3x-4=(x+4)(x-1)

x^3-2x^2+x=x(x^2-2x+1)=x(2-1)(x-1)

x^2+2x-3=(x+3)(x-1)

Therefore

HCF =(x-1)

LCM = x(x+3)(x-1)(x+4)(x-1)

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10)\ \ x^2-16, x^2-3x-28 \ \ and\ \ x^2-11x+28

We have

x^2-16=(x-4)(x+4)

x^2-3x-28=(x-7)(x+4)

x^2-11x+28=(x-7)(x-4)

Therefore

HCF = 1

LCM = (x-7)(x-4)(x+4)

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11) \ \ 12x^2-75, 4x^2-20x+25 \ \ and\ \ 6x^2-13x-5

We have

12x^2-75=3(4x^2-25)=3(2x-5)(2x+5)

4x^2-20x+25=(2x-5)(2x-5)

6x^2-13x-5=(2x-5)(3x-1)

Therefore

HCF = (2x-5)

LCM = 3(2x-5)(3x-1)(2x-5)(2x+5)

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12) \ \ 2a^2+a-6, (2a-3)^2 \ \ and\ \ (4a^2-9)

We have

2a^2+a-6= 2a^2+4a-3a-6=2a(a+2)-3(a+2)=(a+2)(2a+3)

(2a-3)^2=(2a-3)(2a-3)

(4a^2-9)=(2a-3)(2a+3)

Therefore

HCF = (2a-3)

LCM = (2a-3)(2a+3)(2a-3)(a+2)

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13) \ \ 4m^4 n (m^4-n^4 ), 6m^3 n^2 (m^2+2mn-3n^2 ), 24m^2 n^3 (m^3+m^2 n+mn^2+n^3)

We have

4m^4 n(m^4-n^4 )= 4m^4 n (m^2-n^2 )(m^2+n^2 )=4m^4 n (m-n)(m+n)(m^2+n^2 )

6m^3 n^2 (m^2+2mn-3n^2)= 6m^3 n^2 (m+3m)(m-n)

24m^2 n^3 (m^3+m^2 n+mn^2+n^3 )=24mn^2 n^3 (m+n)(m^2+n^2)

Therefore

HCF = 2m^2 n

LCM = 24m^4 n^3 (m-n)(m+n)(m+3n)(m^2+n^2)

 

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