The H.C.F of $24 (6 x^{4} - x^{3} - 2 x^{2})$ and $20 (2 x^{6} + 3 x^{5} + x^{4})$ is:

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4 x^{2}

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4 x^{2} (2 x + 1)

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2 x (2 x + 1)

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Solution

The correct option is D $4 x^{2} (2 x + 1)$
$24 (6 x^{4} - x^{3} - 2 x^{2})$
$= 2^{3} \times 3 \times x^{2} (6 x^{2} - x - 2)$
$= 2^{3} \times 3 \times x^{2} (3 x - 2) (2 x + 1)$
$= 20 (2 x^{6} + 3 x^{5} + x^{4}) = 20 x^{4} (2 x^{2} + 3 x + 1)$
$= 2^{2} \times 5 \times x^{4} (x + 1) (2 x + 1)$
HCF of the two expression is
$= 2^{2} x^{2} (2 x + 1) = 4 x^{2} (2 x + 1)$

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