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Lecture 1

21a Lecture 1: Class 8 -Simplification of Algebraic Equations- Exercise 21A


Department
Mathematics
Course Code
Mathematics 21a
Professor
math
Lecture
1

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Class 8: Simplification of Algebraic Equations- Exercise 21A
Find HCF LCM of the following monomials:
1) \ \ ab^2 \ \ and\ \ a^2 b
HCF = ab
LCM = a^2 b^2
\\
2) \ \ a^3 b^2 \ \ and\ \ a^2 b^4
HCM = a^2 b^2
LCM = a^2 b^4
\\
3) \ \ 4x^2 y^3 \ \ and\ \ 6xy^4
HCF = 2xy^3
LCM = 12x^2 y^4
\\
4) \ \ 6abc \ \ and\ \ 9bc^2 d
HCM = 3bc
LCM = 18abc^2 d
\\
5) \ \ 2m^2 n^3, \ \ 3mn^2 \ \ and\ \ 4m^3 n
HCM = mn
LCM = 12m^3 n^3
\\
6) \ \ 5x^3 y^2, 10x^2 z^2 \ \ and\ \ 15y^3 z^3
HCF = 5x^2
LCM = 30x^3 y^3 z^3
\\ m34x
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
\\
Find HCF and LCM of following polynomials
1) \ \ x^2-a^2 \ \ and\ \ x^2-ax
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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)
\\
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)
\\
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)
\\
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)
\\
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)
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