The HCF of $10 (x^{2} + x - 20), 15 (x^{2} - 3 x - 4)$ , and $20 (x^{2} + 2 x + 1)$ is

5(x-4)

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5(x+1)

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5(x+1)(x-4)

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Solution

The correct option is A 5
The HCF of 10, 15, and 20 is 5.
$x^{2} + x - 20 = (x + 5) (x - 4)$
$x^{2} - 3 x - 4 = (x - 4) (x + 1)$
$x^{2} + 2 x + 1 = (x + 1)^{2}$
$H C F = 5$

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