$The HCF of 6 (4 x^{2} - 4 x + 1) and 9 (2 x^{2} - 7 x + 3) is :$

3 (2 x - 1)

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9 (2 x - 1)

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3 (2 x - 1)^{2}

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18 (2 x - 1)^{2} (x - 3)

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Solution

The correct option is A $3 (2 x - 1)$
Using the identity
$(a - b)^{2} = a^{2} - 2 a b + b^{2}$ ,
We can write
$(4 x^{2} - 4 x + 1) = (2 x - 1)^{2}$
Hence, $6 (4 x^{2} - 4 x + 1) = 2 \times 3 \times (2 x - 1)^{2}$ .

Also, $(2 x^{2} - 7 x + 3) = (2 x^{2} - 6 x - x + 3) = 2 x (x - 3) - 1 (x - 3) = (2 x - 1) (x - 3)$ .

Hence, $9 (2 x^{2} - 7 x + 3) = 3^{2} (2 x - 1) (x - 3)$ .

So, the common irreducible factors occuring in $6 (4 x^{2} - 4 x + 1) a n d 9 (2 x^{2} - 7 x + 3)$ are 3 and (2x-1).

The least exponents of these factors respectively are 1 and 1.

Hence, the HCF = $3^{1} \times (2 x - 1)^{1} = 3 (2 x - 1)$ .

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