The square root of $4 a^{2} + 8 a^{3} + 8 a^{2} + 4 a + 1$ using division method.

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Solution

$\div$ $2 a^{2} + 2 a + 1$ <= Quotient
$2 a^{2}$
$4 a^{4} + 8 a^{3} + 8 a^{2} + 4 a + 1$
$4 a^{4}$
$4 a^{2} + 2 a$
$8 a^{3} + 8 a^{2}$
$8 a^{3} + 4 a^{2}$
$4 a^{2} + 4 a + 1$
$4 a^{2} + 4 a + 1$
$4 a^{2} + 4 a + 1$
Remainder $\Rightarrow 0$
The square root of $4 a^{2} + 8 a^{3} + 8 a^{2} + 4 a + 1$ is $∣ ∣ 2 a^{2} + 2 a + 1 ∣ ∣$

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