Question

6. Find the HCF of : (a) 21xy³ and 24x²y² (b) 18ab and 36abc
(c) 4p³q²r , -12pr² , 16p²q²r²r²​

Answer 1

(a) 3xy^2(7y + 8x)
(b) 18ab(1 + 2c)
(c) 4pr(p^3q^2r - 3r + 4pq^2r)

Answer 2

Explanation:

➤ Solution :-

a)

Given monomials are 21 xy³ and 24x²y²

21 xy³ = 3×7×x×y×y×y

24x²y² = 3×8×x×x×y×y

HCF of 21xy³ and 24x²y²

=> 3×x×y×y

=> 3xy²

HCF of 21xy³ and 24x²y² = 3xy²

b)

Given monomials are 18 ab and 36abc

18 ab = 18×a×b

36abc = 2×18×a×b×c

HCF of 18 ab and 36abc

=> 18×a×b

=> 18ab

HCF of 18 ab and 36abc = 18ab

c)

Given monomials are 4p³q²r , -12pqr² and 16p²q²r²

4p³q²r = 4×p×p×p×q×q×r

-12pqr² = -3×4×p×q×r×r

16p²q²r² = 4×4×p×p×q×q×r×r

HCF of 4p³q²r , -12pqr² and 16p²q²r²

=> 4×p×q×r

=> 4pqr

HCF of 4p³q²r , -12pqr² and 16p²q²r² = 4pqr

Used Concept :-

→ The HCF of two or more numbers is the highest Common factor

→ The product of least common factors is the HCF of the numbers