Find the H.C.F. of $(a^{3} + 4 a^{2} + 4 a + 1)$ and $(a^{4} + 3 a^{3} + 2 a^{2} + 3 a + 1)$ by using the division method.

(a^{2} + 3 a + 1)

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(a^{2} + 3 a - 1)

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(a^{2} - 3 a + 1)

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(a^{2} - 3 a - 1)

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Solution

The correct option is A $(a^{2} + 3 a + 1)$
$a^{3} + 4 a^{2} + 4 a + 1) a^{4} + 3 a^{3} + 2 a^{2} + 3 a + 1 (a - 1 a^{4} + 4 a^{3} + 4 a^{2} + a + - - - - - --------------------- - - a^{3} - 2 a^{2} + 2 a + 1 - a^{3} - 4 a^{2} - 4 a - 1 + + + + - ------------------------ - 2 a^{2} + 6 a + 2$
$2 a^{2} + 6 a + 2 = 2 (a^{2} + 3 a + 1)$

$a^{2} + 3 a + 1) a^{3} + 4 a^{2} + 4 a + 1 (a - 1 a^{3} + 3 a^{2} + a - - - - ------------------- - a^{2} + 3 a + 1 a^{2} + 3 a + 1 - - - - ------------------- - 0$

So, H.C.F. of $(a^{3} + 4 a^{2} + 4 a + 1)$ and $(a^{4} + 3 a^{3} + 2 a^{2} + 3 a + 1)$ is $(a^{2} + 3 a + 1) .$

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