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Question

The HCF of 21x(x21) and 3x3(x+1) is:

A
3x(x+1)
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B
3x(x1)
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C
21x(x+1)
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D
21x(x1)
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Solution

The correct option is A 3x(x+1)
21x(x21)=3× 7× x× (x+1)(x1).
3x3(x+1)=3× x3× (x+1).
Hence, the common irreducible factors between 21x(x21) and 3x3(x+1) are 3,x and (x+1).
The least exponents of these factors are respectively 1,1 and 1.
Hence, the HCF = 31× x1× (x+1)1=3x(x+1).

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