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Question

The LCM of 6(4x24x+1) and 9(2x27x+3) is:

A
(2x1)(x3)
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B
18(2x1)2(x3)
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C
18(2x1)(x3)
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D
9(2x1)(x3)
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Solution

The correct option is B 18(2x1)2(x3)
Using the identity

(ab)2=a22ab+b2,

We can write

(4x24x+1)=(2x1)2

Hence, 6(4x24x+1)=2× 3× (2x1)2.

Also, (2x27x+3)=(2x26xx+3)=2x(x3)1(x3)=(2x1)(x3).

Hence, 9(2x27x+3)=32(2x1)(x3).

So, the irreducible factors occuring in 6(4x24x+1) and 9(2x27x+3) are 2,3,(2x1)and(x3)

The greatest exponents of these factors respectively are 1,2,2 and 1.

Hence, the LCM = 21× 32× (2x1)2× (x3)1=18(2x1)2(x3).

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