Find H.C.F. and L.C.M.:
Q1. $4x^{2}yz^{3}$ and $6x^{3}y^{2}z^{2}$
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Given: $4x^{2}yz^{3} = (2)(2)(x)(x)(y)(z)(z)(z)$
Given: $6x^{3}y^{2}z^{2} = (2)(3)(x)(x)(x)(y)(y)(z)(z)$
Common: $2x^{2}yz^{2}$
And remaining: $(2)3xyz$
$\therefore \text{H.C.F.} = 2x^{2}yz^{2}$ Ans
$\therefore \text{L.C.M.} = 2x^{2}yz^{2}(6xyz) = 12x^{3}y^{2}z^{3}$ Ans
Given: $6x^{3}y^{2}z^{2} = (2)(3)(x)(x)(x)(y)(y)(z)(z)$
Common: $2x^{2}yz^{2}$
And remaining: $(2)3xyz$
$\therefore \text{H.C.F.} = 2x^{2}yz^{2}$ Ans
$\therefore \text{L.C.M.} = 2x^{2}yz^{2}(6xyz) = 12x^{3}y^{2}z^{3}$ Ans
Q2. $3a^{2}b^{2}c, 6a^{3}b$ and $a^{3}b^{2}c^{3}$
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Given: $3a^{2}b^{2}c = (3)(a)(a)(b)(b)(c)$
Given: $6a^{3}b = (2)(3)(a)(a)(a)(b)$
Given: $a^{3}b^{2}c^{3} = (a)(a)(a)(b)(b)(c)(c)(c)$
Common: $a^{2}b$
And remaining: $(2)3abc^{3} = 6abc^{3}$
$\therefore \text{H.C.F.} = a^{2}b$ Ans
$\therefore \text{L.C.M.} = a^{2}b(6abc^{3}) = 6a^{3}b^{2}c^{3}$ Ans
Given: $6a^{3}b = (2)(3)(a)(a)(a)(b)$
Given: $a^{3}b^{2}c^{3} = (a)(a)(a)(b)(b)(c)(c)(c)$
Common: $a^{2}b$
And remaining: $(2)3abc^{3} = 6abc^{3}$
$\therefore \text{H.C.F.} = a^{2}b$ Ans
$\therefore \text{L.C.M.} = a^{2}b(6abc^{3}) = 6a^{3}b^{2}c^{3}$ Ans
Q3. $2xy^{2}, 3x^{2}y^{3}$ and $4x^{3}y$
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Given: $2xy^{2} = (2)(x)(y)(y)$
Given: $3x^{2}y^{3} = (3)(x)(x)(y)(y)(y)$
Given: $4x^{3}y = (2)(2)(x)(x)(x)(y)$
Common: $xy$
And remaining: $(2)(2)(3)x^{2}y^{2} = 12x^{2}y^{2}$
$\therefore \text{H.C.F.} = xy$ Ans
$\therefore \text{L.C.M.} = xy(12x^{2}y^{2}) = 12x^{3}y^{3}$ Ans
Given: $3x^{2}y^{3} = (3)(x)(x)(y)(y)(y)$
Given: $4x^{3}y = (2)(2)(x)(x)(x)(y)$
Common: $xy$
And remaining: $(2)(2)(3)x^{2}y^{2} = 12x^{2}y^{2}$
$\therefore \text{H.C.F.} = xy$ Ans
$\therefore \text{L.C.M.} = xy(12x^{2}y^{2}) = 12x^{3}y^{3}$ Ans
Q4. $x^{2} - y^{2}$ and $x^{2} - xy$
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Given: $x^{2} - y^{2} = (x + y)(x - y)$
Given: $x^{2} - xy = x(x - y)$
Common: $x - y$
And remaining: $x(x + y)$
$\therefore \text{H.C.F.} = x - y$ Ans
$\therefore \text{L.C.M.} = (x - y)x(x + y) = x(x + y)(x - y)$ Ans
Given: $x^{2} - xy = x(x - y)$
Common: $x - y$
And remaining: $x(x + y)$
$\therefore \text{H.C.F.} = x - y$ Ans
$\therefore \text{L.C.M.} = (x - y)x(x + y) = x(x + y)(x - y)$ Ans
Q5. $9x^{2} - 4y^{2}$ and $6x^{2} + 4xy$
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Given: $9x^{2} - 4y^{2} = (3x)^{2} - (2y)^{2} = (3x + 2y)(3x - 2y)$
Given: $6x^{2} + 4xy = 2x(3x - 2y)$
Common: $3x - 2y$
And remaining: $2x(3x + 2y)$
$\therefore \text{H.C.F.} = 3x + 2y$ Ans
$\therefore \text{L.C.M.} = (3x - 2y) 2x(3x + 2y) = 2x(3x + 2y)(3x - 2y)$ Ans
Given: $6x^{2} + 4xy = 2x(3x - 2y)$
Common: $3x - 2y$
And remaining: $2x(3x + 2y)$
$\therefore \text{H.C.F.} = 3x + 2y$ Ans
$\therefore \text{L.C.M.} = (3x - 2y) 2x(3x + 2y) = 2x(3x + 2y)(3x - 2y)$ Ans
Q6. $a^{2}b^{2} - b^{4}, ab^{2} + b^{3}$ and $ab - b^{2}$
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Given: $a^{2}b^{2} - b^{4} = b^{2}(a^{2} - b^{2}) = (b)(b)(a + b)(a - b)$
Given: $ab^{2} + b^{3} = b^{2}(a + b) = (b)(b)(a + b)$
Given: $ab - b^{2} = b(a - b)$
Common: $b$
And remaining: $b(a + b)(a - b)$
$\therefore \text{H.C.F.} = b$ Ans
$\therefore \text{L.C.M.} = b(b)(a + b)(a - b) = b^{2}(a^{2} - b^{2})$ Ans
Given: $ab^{2} + b^{3} = b^{2}(a + b) = (b)(b)(a + b)$
Given: $ab - b^{2} = b(a - b)$
Common: $b$
And remaining: $b(a + b)(a - b)$
$\therefore \text{H.C.F.} = b$ Ans
$\therefore \text{L.C.M.} = b(b)(a + b)(a - b) = b^{2}(a^{2} - b^{2})$ Ans
Q7. $a^{3} - 36a$ and $a^{3} + 2a^{2} - 48a$
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Given: $a^{3} - 36a = a(a^{2} - 36) = a(a^{2} - 6^{2}) = a(a + 6)(a - 6)$
Given: $a^{3} + 2a^{2} - 48a = a^{3} + 8a^{2} - 6a^{2} - 48a$
$\Rightarrow a^{3} + 8a^{2} - 6a^{2} - 48a = a^{2}(a + 8) - 6a(a + 8)$
$\Rightarrow a^{2}(a + 8) - 6a^{2}(a + 8) = (a + 8)(a^{2} - 6a)$
$\Rightarrow (a + 8)(a^{2} - 6) = (a + 8)a(a - 6) = a(a + 8)(a - 6)$
Common: $a(a - 6)$
And remaining: $(a + 6)(a + 8)$
$\therefore \text{H.C.F.} = a(a - 6)$ Ans
$\therefore \text{L.C.M.} = a(a + 6)(a - 6)(a + 8) = a(a^{2} - 6^{2})(a + 8)$ Ans
Given: $a^{3} + 2a^{2} - 48a = a^{3} + 8a^{2} - 6a^{2} - 48a$
$\Rightarrow a^{3} + 8a^{2} - 6a^{2} - 48a = a^{2}(a + 8) - 6a(a + 8)$
$\Rightarrow a^{2}(a + 8) - 6a^{2}(a + 8) = (a + 8)(a^{2} - 6a)$
$\Rightarrow (a + 8)(a^{2} - 6) = (a + 8)a(a - 6) = a(a + 8)(a - 6)$
Common: $a(a - 6)$
And remaining: $(a + 6)(a + 8)$
$\therefore \text{H.C.F.} = a(a - 6)$ Ans
$\therefore \text{L.C.M.} = a(a + 6)(a - 6)(a + 8) = a(a^{2} - 6^{2})(a + 8)$ Ans
Q8. $x^{2} - 7x + 10$ and $x^{2} - 4x + 4$
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Given: $x^{2} - 7x + 10 = x^{2} - 5x - 2x + 10$
$\Rightarrow x^{2} - 5x - 2x + 10 = x(x - 5) - 2(x - 5)$
$\Rightarrow x(x - 5) - 2(x - 5) = (x - 5)(x - 2)$
Given: $x^{2} - 4x + 4 = x^{2} - 2x - 2x + 4^{2}$
$\Rightarrow x^{2} - 2x - 2x + 4^{2} = x(x - 2) - 2(x - 2)$
$\Rightarrow x(x - 2) - 2(x - 2) = (x - 2)(x - 2)$
Common: $x - 2$
And remaining: $(x - 5)(x - 2)$
$\therefore \text{H.C.F.} = x - 2$ Ans
$\therefore \text{L.C.M.} = (x - 2)(x - 5)(x - 2) = (x - 5)(x - 2)^{2}$ Ans
$\Rightarrow x^{2} - 5x - 2x + 10 = x(x - 5) - 2(x - 5)$
$\Rightarrow x(x - 5) - 2(x - 5) = (x - 5)(x - 2)$
Given: $x^{2} - 4x + 4 = x^{2} - 2x - 2x + 4^{2}$
$\Rightarrow x^{2} - 2x - 2x + 4^{2} = x(x - 2) - 2(x - 2)$
$\Rightarrow x(x - 2) - 2(x - 2) = (x - 2)(x - 2)$
Common: $x - 2$
And remaining: $(x - 5)(x - 2)$
$\therefore \text{H.C.F.} = x - 2$ Ans
$\therefore \text{L.C.M.} = (x - 2)(x - 5)(x - 2) = (x - 5)(x - 2)^{2}$ Ans
Q9. $a^{2} + a - 6, a^{2} + 2a - 8$ and $2a^{2} - 5a + 2$
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Given: $a^{2} + a - 6 = a^{2} + 3a - 2a - 6$
$\Rightarrow a^{2} + 3a - 2a - 6 = a(a + 3) - 2(a + 3)$
$\Rightarrow a(a + 3) - 2(a + 3) = (a + 3)(a - 2)$
Given: $a^{2} + 2a - 8 = a^{2} - 4a + 2a - 8$
$\Rightarrow a^{2} + 4a - 2a - 8 = a(a + 4) - 2(a + 4)$
$\Rightarrow a(a + 4) - 2(a + 4) = (a + 4)(a - 2)$
Given: $2a^{2} - 5a + 2 = 2a^{2} - 4a - a + 2$
$\Rightarrow 2a^{2} - 4a - a + 2 = 2a(a - 2) - 1(a - 2)$
$\Rightarrow 2a(a - 2) - 1(a - 2) = (a - 2)(2a - 1)$
Common: $a - 2$
And remaining: $(a + 3)(a + 4)(2a - 1)$
$\therefore \text{H.C.F.} = a - 2$ Ans
$\therefore \text{L.C.M.} = (a - 2)(a + 3)(a + 4)(2a - 1)$ Ans
$\Rightarrow a^{2} + 3a - 2a - 6 = a(a + 3) - 2(a + 3)$
$\Rightarrow a(a + 3) - 2(a + 3) = (a + 3)(a - 2)$
Given: $a^{2} + 2a - 8 = a^{2} - 4a + 2a - 8$
$\Rightarrow a^{2} + 4a - 2a - 8 = a(a + 4) - 2(a + 4)$
$\Rightarrow a(a + 4) - 2(a + 4) = (a + 4)(a - 2)$
Given: $2a^{2} - 5a + 2 = 2a^{2} - 4a - a + 2$
$\Rightarrow 2a^{2} - 4a - a + 2 = 2a(a - 2) - 1(a - 2)$
$\Rightarrow 2a(a - 2) - 1(a - 2) = (a - 2)(2a - 1)$
Common: $a - 2$
And remaining: $(a + 3)(a + 4)(2a - 1)$
$\therefore \text{H.C.F.} = a - 2$ Ans
$\therefore \text{L.C.M.} = (a - 2)(a + 3)(a + 4)(2a - 1)$ Ans
Q10. $a^{2} + 3a, a^{2} + 4a + 3$ and $a^{2} + 3a + 2$
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Given: $a^{2} + 3a = a(a + 3)$
Given: $a^{2} + 4a + 3 = a^{2} + 3a + a + 3$
$\Rightarrow a^{2} + 3a + a + 3 = a(a + 3) + 1(a + 3)$
$\Rightarrow a(a + 3) + 1(a + 3) = (a + 3)(a + 1)$
Given: $a^{2} + 3a + 2 = a^{2} + 2a + a + 2$
$\Rightarrow a^{2} + 2a + a + 2 = a(a + 2) + 1(a + 2)$
$\Rightarrow a(a + 2) + 1(a + 2) = (a + 2)(a + 1)$
Common: 1 (when none)
And remaining: $a(a + 3)(a + 1)(a + 2)$
$\therefore \text{H.C.F.} = 1$ Ans
$\therefore \text{L.C.M.} = a(a + 3)(a + 1)(a + 2)$ Ans
Given: $a^{2} + 4a + 3 = a^{2} + 3a + a + 3$
$\Rightarrow a^{2} + 3a + a + 3 = a(a + 3) + 1(a + 3)$
$\Rightarrow a(a + 3) + 1(a + 3) = (a + 3)(a + 1)$
Given: $a^{2} + 3a + 2 = a^{2} + 2a + a + 2$
$\Rightarrow a^{2} + 2a + a + 2 = a(a + 2) + 1(a + 2)$
$\Rightarrow a(a + 2) + 1(a + 2) = (a + 2)(a + 1)$
Common: 1 (when none)
And remaining: $a(a + 3)(a + 1)(a + 2)$
$\therefore \text{H.C.F.} = 1$ Ans
$\therefore \text{L.C.M.} = a(a + 3)(a + 1)(a + 2)$ Ans
Q11. $x^{2} - 25, x^{2} - 2x - 35$ and $x^{2} - 12x + 35$
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Given: $x^{2} - 25 = x^{2} - 5^{2} = (x + 5)(x - 5)$
Given: $x^{2} - 2x - 35 = x^{2} - 7x + 5x - 35$
$\Rightarrow x^{2} - 7x + 5x - 35 = x(x - 7) + 5(x - 7)$
$\Rightarrow x(x - 7) + 5(x - 7) = (x - 7)(x + 5)$
Given: $x^{2} - 12x + 35 = x^{2} - 7x - 5x + 35$
$\Rightarrow x^{2} - 7x - 5x + 35 = x(x - 7) - 5(x - 7)$
$\Rightarrow x(x - 7) - 5(x - 7) = (x - 7)(x - 5)$
Common: 1
And remaining: $(x + 5)(x - 5)(x - 7)$
$\therefore \text{H.C.F.} = 1$ Ans
$\therefore \text{L.C.M.} = (x + 5)(x - 5)(x - 7)$ Ans
Given: $x^{2} - 2x - 35 = x^{2} - 7x + 5x - 35$
$\Rightarrow x^{2} - 7x + 5x - 35 = x(x - 7) + 5(x - 7)$
$\Rightarrow x(x - 7) + 5(x - 7) = (x - 7)(x + 5)$
Given: $x^{2} - 12x + 35 = x^{2} - 7x - 5x + 35$
$\Rightarrow x^{2} - 7x - 5x + 35 = x(x - 7) - 5(x - 7)$
$\Rightarrow x(x - 7) - 5(x - 7) = (x - 7)(x - 5)$
Common: 1
And remaining: $(x + 5)(x - 5)(x - 7)$
$\therefore \text{H.C.F.} = 1$ Ans
$\therefore \text{L.C.M.} = (x + 5)(x - 5)(x - 7)$ Ans
Q12. $2a^{2} - a - 3, (2a - 3)^{2}$ and $4a^{2} - 9$
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Given: $2a^{2} - a - 3 = 2a^{2} - 3a + 2a - 3$
$\Rightarrow 2a^{2} - 3a + 2a - 3 = a(2a - 3) + 1(2a - 3)$
$\Rightarrow a(2a - 3) + 1(2a - 3) = (2a - 3)(a + 1)$
Given: $(2a - 3)^{2} = (2a - 3)(2a - 3)$
Given: $4a^{2} - 9 = 2a^{2} - 3^{2} = (2a + 3)(2a - 3)$
Common: $2a - 3$
And remaining: $(a + 1)(2a - 3)(2a + 3)$
$\therefore \text{H.C.F.} = 2a - 3$ Ans
$\therefore \text{L.C.M.} = (2a - 3)(a + 1)(2a - 3)(2a + 3) = (2a - 3)^{2}(a + 1)(2a + 3)$ Ans
$\Rightarrow 2a^{2} - 3a + 2a - 3 = a(2a - 3) + 1(2a - 3)$
$\Rightarrow a(2a - 3) + 1(2a - 3) = (2a - 3)(a + 1)$
Given: $(2a - 3)^{2} = (2a - 3)(2a - 3)$
Given: $4a^{2} - 9 = 2a^{2} - 3^{2} = (2a + 3)(2a - 3)$
Common: $2a - 3$
And remaining: $(a + 1)(2a - 3)(2a + 3)$
$\therefore \text{H.C.F.} = 2a - 3$ Ans
$\therefore \text{L.C.M.} = (2a - 3)(a + 1)(2a - 3)(2a + 3) = (2a - 3)^{2}(a + 1)(2a + 3)$ Ans
Q13. $x^{3}y - xy^{3}$ and $x^{3}y^{2} + x^{2}y^{3}$
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Given: $x^{3}y - xy^{3} = xy(x^{2} - y^{2}) = xy(x + y)(x - y)$
Given: $x^{3}y^{2} + x^{2}y^{3} = x^{2}y^{2}(x + y)$
Common: $x + y$
And remaining: $x^{2}y^{2}(x - y)$
$\therefore \text{H.C.F.} = x + y$ Ans
$\therefore \text{L.C.M.} = (x + y)x^{2}y^{2}(x - y) = x^{2}y^{2}(x^{2} - y^{2})$ Ans
Given: $x^{3}y^{2} + x^{2}y^{3} = x^{2}y^{2}(x + y)$
Common: $x + y$
And remaining: $x^{2}y^{2}(x - y)$
$\therefore \text{H.C.F.} = x + y$ Ans
$\therefore \text{L.C.M.} = (x + y)x^{2}y^{2}(x - y) = x^{2}y^{2}(x^{2} - y^{2})$ Ans
Q14. $x^{3} - 5x^{2} + 6x$ and $x^{3} + 4x^{2} - 12x$
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Given: $x^{3} - 5x^{2} + 6x = x^{3} - 3x^{2} - 2x^{2} + 6x$
$\Rightarrow x^{3} - 3x^{2} - 2x^{2} + 6x = x^{2}(x - 3) - 2x(x - 3)$
$\Rightarrow x^{2}(x - 3) - 2x(x - 3) = (x - 3)(x^{2} - 2x)$
$\Rightarrow (x - 3)(x^{2} - 2x) = (x - 3)x(x - 2) = x(x - 3)(x - 2)$
Given: $x^{3} + 4x^{2} - 12x = x^{3} + 6x^{2} - 2x^{2} - 12x$
$\Rightarrow x^{3} + 6x^{2} - 2x^{2} - 12x = x^{2}(x + 6) - 2x(x + 6)$
$\Rightarrow x^{2}(x + 6) - 2x(x + 6) = (x + 6)(x^{2} - 2x)$
$\Rightarrow (x + 6)(x^{2} - 2x) = (x + 6)x(x - 2) = x(x + 6)(x - 2)$
Common: $x - 2$
And remaining: $x(x - 3)(x + 6)$
$\therefore \text{H.C.F.} = x - 2$ Ans
$\therefore \text{L.C.M.} = (x - 2)x(x - 3)(x + 6) = x(x - 2)(x - 3)(x + 6)$ Ans
$\Rightarrow x^{3} - 3x^{2} - 2x^{2} + 6x = x^{2}(x - 3) - 2x(x - 3)$
$\Rightarrow x^{2}(x - 3) - 2x(x - 3) = (x - 3)(x^{2} - 2x)$
$\Rightarrow (x - 3)(x^{2} - 2x) = (x - 3)x(x - 2) = x(x - 3)(x - 2)$
Given: $x^{3} + 4x^{2} - 12x = x^{3} + 6x^{2} - 2x^{2} - 12x$
$\Rightarrow x^{3} + 6x^{2} - 2x^{2} - 12x = x^{2}(x + 6) - 2x(x + 6)$
$\Rightarrow x^{2}(x + 6) - 2x(x + 6) = (x + 6)(x^{2} - 2x)$
$\Rightarrow (x + 6)(x^{2} - 2x) = (x + 6)x(x - 2) = x(x + 6)(x - 2)$
Common: $x - 2$
And remaining: $x(x - 3)(x + 6)$
$\therefore \text{H.C.F.} = x - 2$ Ans
$\therefore \text{L.C.M.} = (x - 2)x(x - 3)(x + 6) = x(x - 2)(x - 3)(x + 6)$ Ans
Q15. $a^{2} - 7a + 12, 3a^{2} - 6a - 9$ and $2a^{3} - 6a^{2} - 8a$
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Given: $a^{2} - 7a + 12 = a^{2} - 4a - 3a + 12$
$\Rightarrow a^{2} - 4a - 3a + 12 = a(a - 4) - 3(a - 4)$
$\Rightarrow a(a - 4) - 3(a - 4) = (a - 4)(a - 3)$
Given: $3a^{2} - 6a - 9 = 3a^{2} - 9a + 3a - 9$
$\Rightarrow 3a^{2} - 9a + 3a - 9 = 3a(a - 3) + 3(a - 3)$
$\Rightarrow 3a(a - 3) + 3(a - 3) = (a - 3)(3a + 3)$
$\Rightarrow (a - 3)(3a + 3) = (a - 3)3(a + 1) = 3(a - 3)(a + 1)$
Given: $2a^{3} - 6a^{2} - 8a = 2a^{3} - 8a^{2} + 2a^{2} - 8a$
$\Rightarrow 2a^{3} - 8a^{2} + 2a^{2} - 8a = 2a^{2}(a - 4) + 2a(a - 4)$
$\Rightarrow 2a^{2}(a - 4) + 2a(a - 4) = (a - 4)(2a^{2} + 2a)$
$\Rightarrow (a - 4)(2a^{2} + 2a) = (a - 4)2a(a + 1) = 2a(a - 4)(a + 1)$
Common: 1
And remaining: $(a - 4)(a - 3)3(a + 1)2a$
$\therefore \text{H.C.F.} = 1$ Ans
$\therefore \text{L.C.M.} = 6a(a - 4)(a - 3)(a + 1)$ Ans
$\Rightarrow a^{2} - 4a - 3a + 12 = a(a - 4) - 3(a - 4)$
$\Rightarrow a(a - 4) - 3(a - 4) = (a - 4)(a - 3)$
Given: $3a^{2} - 6a - 9 = 3a^{2} - 9a + 3a - 9$
$\Rightarrow 3a^{2} - 9a + 3a - 9 = 3a(a - 3) + 3(a - 3)$
$\Rightarrow 3a(a - 3) + 3(a - 3) = (a - 3)(3a + 3)$
$\Rightarrow (a - 3)(3a + 3) = (a - 3)3(a + 1) = 3(a - 3)(a + 1)$
Given: $2a^{3} - 6a^{2} - 8a = 2a^{3} - 8a^{2} + 2a^{2} - 8a$
$\Rightarrow 2a^{3} - 8a^{2} + 2a^{2} - 8a = 2a^{2}(a - 4) + 2a(a - 4)$
$\Rightarrow 2a^{2}(a - 4) + 2a(a - 4) = (a - 4)(2a^{2} + 2a)$
$\Rightarrow (a - 4)(2a^{2} + 2a) = (a - 4)2a(a + 1) = 2a(a - 4)(a + 1)$
Common: 1
And remaining: $(a - 4)(a - 3)3(a + 1)2a$
$\therefore \text{H.C.F.} = 1$ Ans
$\therefore \text{L.C.M.} = 6a(a - 4)(a - 3)(a + 1)$ Ans