Factorise : $16 x^{4} - 1$

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Solution

Given : $16 x^{4} - 1$
We can further write it as
$16 x^{4} - 1$ $= (4 x^{2})^{2} - 1^{2}$
Based on the equation $a^{2} - b^{2} = (a + b) (a - b)$
We get,
$16 x^{4} - 1$ $= (4 x^{2} + 1) (4 x^{2} - 1)$
On further simplification using the formula
$16 x^{4} - 1$ $= (4 x^{2} + 1) ((2 x)^{2} - 1^{2})$
So we get,
$16 x^{4} - 1$ $= (4 x^{2} + 1) (2 x + 1) (2 x - 1)$

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