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

$HCF = 2m^2 n$

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