How do you factor completely $n^{4} + 2 n^{2} w^{2} + w^{4}$ $n^4 + 2n^2w^2+ w^4$ ?

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How do you factor completely $n^{4} + 2 n^{2} w^{2} + w^{4}$ $n^4 + 2n^2w^2+ w^4$ ?

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Answer 1 · 2016-01-02T21:46:06

$n^{4} + 2 n^{2} w^{2} + w^{4}$ $n^4+2n^2w^2+w^4$

$= {(n^{2} + w^{2})}^{2}$ $=(n^2+w^2)^2$

$= {(n - i w)}^{2} {(n + i w)}^{2}$ $=(n-iw)^2(n+iw)^2$

Explanation:

$n^{4} + 2 n^{2} w^{2} + w^{4}$ $n^4+2n^2w^2+w^4$

$= {(n^{2})}^{2} + 2 (n^{2}) (w^{2}) + {(w^{2})}^{2}$ $=(n^2)^2+2(n^2)(w^2)+(w^2)^2$

$= {(n^{2} + w^{2})}^{2}$ $=(n^2+w^2)^2$

In addition, if we allow Complex coefficients then $n^{2} + w^{2}$ $n^2+w^2$ can be treated as a difference (!) of squares, being of the form $a^{2} - b^{2}$ $a^2-b^2$ with $a = n$ $a=n$ and $b = i w$ $b=iw$ .

The difference of squares identity can be written:

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

So we find:

$n^{2} + w^{2} = n^{2} - {(i w)}^{2} = (n - i w) (n + i w)$ $n^2+w^2 = n^2 - (iw)^2 = (n-iw)(n+iw)$

Hence we can write:

$n^{4} + 2 n^{2} w^{2} + w^{4} = {(n - i w)}^{2} {(n + i w)}^{2}$ $n^4+2n^2w^2+w^4 = (n-iw)^2(n+iw)^2$

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How do you factor completely $n^{4} + 2 n^{2} w^{2} + w^{4}$ $n^4 + 2n^2w^2+ w^4$ ?

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Explanation:

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How do you factor completely n^4 + 2n^2w^2+ w^4n4+2n2w2+w4n^4 + 2n^2w^2+ w^4?

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How do you factor completely $n^{4} + 2 n^{2} w^{2} + w^{4}$ $n^4 + 2n^2w^2+ w^4$ ?