Find the factors $a^{4} + 4 a^{2} + 16$

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Solution

In the given expression $a^{4} + 4 a^{2} + 16$ , add and subtract $4 a^{2}$ to make it a perfect square as shown below:

$a^{4} + 4 a^{2} + 16 = (a^{4} + 4 a^{2} + 16 + 4 a^{2}) - 4 a^{2} = (a^{4} + 8 a^{2} + 16) - 4 a^{2} = [(a^{2})^{2} + (4)^{2} + 2 (a^{2}) (4)] - 4 a^{2}$
$= (a^{2} + 4)^{2} - 4 a^{2}$
(using the identity $(a + b)^{2} = a^{2} + b^{2} + 2 a b$ )

We also know the identity $a^{2} - b^{2} = (a + b) (a - b)$ , therefore,

Using the above identity, the expression $(a^{2} + 4)^{2} - 4 a^{2}$ can be factorised as follows:

$(a^{2} + 4)^{2} - 4 a^{2} = (a^{2} + 4)^{2} - (2 a)^{2} = (a^{2} + 4 + 2 a) (a^{2} + 4 - 2 a)$

Hence, $a^{4} + 4 a^{2} + 16 = (a^{2} + 4 + 2 a) (a^{2} + 4 - 2 a)$

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