Class 8, Exercise 21A, Simplification of Algebraic Fractions

Class 8: Simplification of Algebraic Equations- Exercise 21A

Date: July 24, 2016Author: ICSE CBSE ISC Board Mathematics Portal for Students 0 Comments

Find HCF LCM of the following monomials:

Question 1: $ab^2$ and $a^2 b$

Answer:

HCF $= ab$ LCM $= a^2 b^2$

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Question 2: $a^3 b^2$ and $a^2 b^4$

Answer:

HCF $= a^2 b^2$ LCM $= a^2 b^4$

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Question 3: $4x^2 y^3$ and $6xy^4$

Answer:

HCF $= 2xy^3$ LCM $= 12x^2 y^4$

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Question 4: $6abc$ and $9bc^2 d$

Answer:

HCF $= 3bc$ LCM $= 18abc^2 d$

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Question 5: $2m^2 n^3, 3mn^2$ and $4m^3 n$

Answer:

HCF $= mn$ LCM $= 12m^3 n^3$

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Question 6: $5x^3 y^2, 10x^2 z^2$ and $15y^3 z^3$

Answer:

HCF $= 5x^2$ LCM $= 30x^3 y^3 z^3$

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Question 7: $6x^2 y^2 z^4, 9x^4 y^5 z$ and $12xy^2 z^3$

Answer:

HCF $= 3xy^2 z$ LCM $= 36x^4 y^5 z^4$

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Find the HCF and LCM of following polynomials:

Question 8: $x^2-a^2$ and $x^2-ax$

Answer:

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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Question 9: $9a^2-16b^2$ and $6a^2+8ab$

Answer:

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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Question 10: $x^3-16x$ and $x^3+2x^2-24x$

Answer:

We have

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

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

Therefore HCF $= x(x-4)$ LCM $= x(x-4)(x+6)(x+4)$

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Question 11: $x^2+x-12$ and $x^2-6x+9$

Answer:

We have

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

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

Therefore HCF $= (x-3)$ LCM $= (x-3)(x-3)(x+4)$

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Question 12: $x^2-36$ and $2x^2-15x+18$

Answer:

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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Question 13: $6a^2 b (a^2-b^2 )$ and $4ab^2 (a+b)^2$

Answer:

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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Question 14: $xy+y^2, xy-y^2$ and $x^2 y^2-y^4$

Answer:

We have

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

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

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

Therefore HCF $= y$ LCM $= y^2 (x-y)(x+y)$

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Question 15: $x^2+3x, x^2+5x+6$ and $x^2+4x+3$

Answer:

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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Question 16: $x^2+3x-4, x^3-2x^2+x$ and $x^2+2x-3$

Answer:

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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Question 17: $x^2-16, x^2-3x-28$ and $x^2-11x+28$

Answer:

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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Question 18: $12x^2-75, 4x^2-20x+25$ and $6x^2-13x-5$

Answer:

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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Question 19: $2a^2+a-6, (2a-3)^2$ and $(4a^2-9)$

Answer:

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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Question 20: $4m^4 n (m^4-n^4 ), 6m^3 n^2 (m^2+2mn-3n^2 ), 24m^2 n^3 (m^3+m^2 n+mn^2+n^3)$

Answer:

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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