Class 8: Simplification of Algebraic Equations- Exercise 21A

$\displaystyle \textbf{Find the H.C.F. and L.C.M. of the following monomials:}$

$\displaystyle \textbf{Question 1: }ab^2\text{ and }a^2b.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{H.C.F.}=ab,\qquad \text{L.C.M.}=a^2b^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }a^3b^2\text{ and }a^2b^4.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{H.C.F.}=a^2b^2,\qquad \text{L.C.M.}=a^3b^4.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }4x^2y^3\text{ and }6xy^4.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{H.C.F. of }4\text{ and }6=2,\qquad \text{L.C.M. of }4\text{ and }6=12.$
$\displaystyle \text{H.C.F.}=2xy^3,\qquad \text{L.C.M.}=12x^2y^4.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }6abc\text{ and }9bc^2d.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{H.C.F. of }6\text{ and }9=3,\qquad \text{L.C.M. of }6\text{ and }9=18.$
$\displaystyle \text{H.C.F.}=3bc,\qquad \text{L.C.M.}=18abc^2d.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }2m^2n^3,\ 3mn^2 \text{ and }4m^3n.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{H.C.F. of }2,3,4=1,\qquad \text{L.C.M. of }2,3,4=12.$
$\displaystyle \text{H.C.F.}=mn,\qquad \text{L.C.M.}=12m^3n^3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }5x^3y^2,\ 10x^2z^2 \text{ and }15y^3z^3.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{H.C.F. of }5,10,15=5.$
$\displaystyle \text{There is no literal common to all three monomials.}$
$\displaystyle \therefore \text{ H.C.F.}=5.$
$\displaystyle \text{L.C.M. of }5,10,15=30.$
$\displaystyle \therefore \text{ L.C.M.}=30x^3y^3z^3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }6x^2y^2z^4,\ 9x^4y^5z \text{ and }12xy^2z^3.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{H.C.F. of }6,9,12=3,\qquad \text{L.C.M. of }6,9,12=36.$
$\displaystyle \text{H.C.F.}=3xy^2z,\qquad \text{L.C.M.}=36x^4y^5z^4.$
$\displaystyle \\$

$\displaystyle \textbf{Find the H.C.F. and L.C.M. of the following polynomials:}$

$\displaystyle \textbf{Question 8: }x^2-a^2\text{ and }x^2-ax.$
$\displaystyle \text{Answer:}$
$\displaystyle x^2-a^2=(x-a)(x+a)$
$\displaystyle x^2-ax=x(x-a)$
$\displaystyle \therefore \text{ H.C.F.}=x-a.$
$\displaystyle \therefore \text{ L.C.M.}=x(x-a)(x+a).$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }9a^2-16b^2 \text{ and }6a^2+8ab.$
$\displaystyle \text{Answer:}$
$\displaystyle 9a^2-16b^2=(3a-4b)(3a+4b)$
$\displaystyle 6a^2+8ab=2a(3a+4b)$
$\displaystyle \therefore \text{ H.C.F.}=3a+4b.$
$\displaystyle \therefore \text{ L.C.M.} =2a(3a-4b)(3a+4b).$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }x^3-16x \text{ and }x^3+2x^2-24x.$
$\displaystyle \text{Answer:}$
$\displaystyle x^3-16x=x(x^2-16)$
$\displaystyle =x(x-4)(x+4)$
$\displaystyle x^3+2x^2-24x=x(x^2+2x-24)$
$\displaystyle =x(x+6)(x-4)$
$\displaystyle \therefore \text{ H.C.F.}=x(x-4).$
$\displaystyle \therefore \text{ L.C.M.} =x(x-4)(x+4)(x+6).$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }x^2+x-12 \text{ and }x^2-6x+9.$
$\displaystyle \text{Answer:}$
$\displaystyle x^2+x-12=(x+4)(x-3)$
$\displaystyle x^2-6x+9=(x-3)^2$
$\displaystyle \therefore \text{ H.C.F.}=x-3.$
$\displaystyle \therefore \text{ L.C.M.}=(x+4)(x-3)^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }x^2-36 \text{ and }2x^2-15x+18.$
$\displaystyle \text{Answer:}$
$\displaystyle x^2-36=(x-6)(x+6)$
$\displaystyle 2x^2-15x+18=2x^2-12x-3x+18$
$\displaystyle =2x(x-6)-3(x-6)$
$\displaystyle =(2x-3)(x-6)$
$\displaystyle \therefore \text{ H.C.F.}=x-6.$
$\displaystyle \therefore \text{ L.C.M.}=(x-6)(x+6)(2x-3).$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }6a^2b(a^2-b^2) \text{ and }4ab^2(a+b)^2.$
$\displaystyle \text{Answer:}$
$\displaystyle 6a^2b(a^2-b^2)=6a^2b(a-b)(a+b)$
$\displaystyle 4ab^2(a+b)^2=4ab^2(a+b)^2$
$\displaystyle \therefore \text{ H.C.F.}=2ab(a+b).$
$\displaystyle \therefore \text{ L.C.M.} =12a^2b^2(a-b)(a+b)^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }xy+y^2,\ xy-y^2 \text{ and }x^2y^2-y^4.$
$\displaystyle \text{Answer:}$
$\displaystyle xy+y^2=y(x+y)$
$\displaystyle xy-y^2=y(x-y)$
$\displaystyle x^2y^2-y^4=y^2(x^2-y^2)$
$\displaystyle =y^2(x-y)(x+y)$
$\displaystyle \therefore \text{ H.C.F.}=y.$
$\displaystyle \therefore \text{ L.C.M.}=y^2(x-y)(x+y).$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }x^2+3x,\ x^2+5x+6 \text{ and }x^2+4x+3.$
$\displaystyle \text{Answer:}$
$\displaystyle x^2+3x=x(x+3)$
$\displaystyle x^2+5x+6=(x+2)(x+3)$
$\displaystyle x^2+4x+3=(x+1)(x+3)$
$\displaystyle \therefore \text{ H.C.F.}=x+3.$
$\displaystyle \therefore \text{ L.C.M.} =x(x+1)(x+2)(x+3).$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }x^2+3x-4,\ x^3-2x^2+x \text{ and }x^2+2x-3.$
$\displaystyle \text{Answer:}$
$\displaystyle x^2+3x-4=(x+4)(x-1)$
$\displaystyle x^3-2x^2+x=x(x^2-2x+1)=x(x-1)^2$
$\displaystyle x^2+2x-3=(x+3)(x-1)$
$\displaystyle \therefore \text{ H.C.F.}=x-1.$
$\displaystyle \therefore \text{ L.C.M.} =x(x+3)(x+4)(x-1)^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }x^2-16,\ x^2-3x-28 \text{ and }x^2-11x+28.$
$\displaystyle \text{Answer:}$
$\displaystyle x^2-16=(x-4)(x+4)$
$\displaystyle x^2-3x-28=(x-7)(x+4)$
$\displaystyle x^2-11x+28=(x-7)(x-4)$
$\displaystyle \therefore \text{ H.C.F.}=1.$
$\displaystyle \therefore \text{ L.C.M.}=(x-7)(x-4)(x+4).$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }12x^2-75,\ 4x^2-20x+25 \text{ and }6x^2-13x-5.$
$\displaystyle \text{Answer:}$
$\displaystyle 12x^2-75=3(4x^2-25)=3(2x-5)(2x+5)$
$\displaystyle 4x^2-20x+25=(2x-5)^2$
$\displaystyle 6x^2-13x-5=6x^2-15x+2x-5$
$\displaystyle =3x(2x-5)+(2x-5)$
$\displaystyle =(3x+1)(2x-5)$
$\displaystyle \therefore \text{ H.C.F.}=2x-5.$
$\displaystyle \therefore \text{ L.C.M.} =3(2x-5)^2(2x+5)(3x+1).$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }2a^2+a-6,\ (2a-3)^2 \text{ and }4a^2-9.$
$\displaystyle \text{Answer:}$
$\displaystyle 2a^2+a-6=2a^2+4a-3a-6$
$\displaystyle =2a(a+2)-3(a+2)$
$\displaystyle =(a+2)(2a-3)$
$\displaystyle (2a-3)^2=(2a-3)(2a-3)$
$\displaystyle 4a^2-9=(2a-3)(2a+3)$
$\displaystyle \therefore \text{ H.C.F.}=2a-3.$
$\displaystyle \therefore \text{ L.C.M.} =(a+2)(2a-3)^2(2a+3).$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: }4m^4n(m^4-n^4),\ 6m^3n^2(m^2+2mn-3n^2),$
$\displaystyle 24m^2n^3(m^3+m^2n+mn^2+n^3).$
$\displaystyle \text{Answer:}$