How do you factor $y^{8 n} - 2 y^{4 n} + 1$ $y^(8n) - 2y^(4n) +1$ ?

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How do you factor $y^{8 n} - 2 y^{4 n} + 1$ $y^(8n) - 2y^(4n) +1$ ?

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Answer 1 · 2016-07-26T16:04:51

Factoring

Explanation:

$f (y) = y^{8 n} - 2 y^{4 n} + 1 = {(y^{4 n} - 1)}^{2} =$ $f(y) = y^(8n) - 2y^(4n) + 1 = (y^(4n) - 1)^2 =$
$= {(y^{2 n} + 1)}^{2} {(y^{2 n} - 1)}^{2} = {(y^{2 n} + 1)}^{2} {(y^{n} + 1)}^{2} {(y^{n} - 1)}^{2}$ $= (y^(2n) + 1)^2(y^(2n) - 1)^2 = (y^(2n) + 1)^2(y^n + 1)^2(y^n -1)^2$

NOTE. We may continue, if required, to factor ${(y^{n} - 1)}^{2}$ $(y^n - 1)^2$

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How do you factor $y^{8 n} - 2 y^{4 n} + 1$ $y^(8n) - 2y^(4n) +1$ ?

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