How do you factor completely: $x^{4} - 1$ $x^4 - 1$ ?

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Answer 1 · 2015-07-17T20:37:27

Use difference of squares identity twice to find:

$x^{4} - 1 = (x - 1) (x + 1) (x^{2} + 1)$ $x^4 - 1 = (x-1)(x+1)(x^2+1)$

The difference of squares identity is:

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

So we find:

$x^{4} - 1$ $x^4 - 1$

$= {(x^{2})}^{2} - 1^{2} = (x^{2} - 1) (x^{2} + 1)$ $= (x^2)^2 - 1^2 = (x^2 - 1)(x^2+1)$

$= (x^{2} - 1^{2}) (x^{2} + 1) = (x - 1) (x + 1) (x^{2} + 1)$ $= (x^2-1^2)(x^2+1) = (x-1)(x+1)(x^2+1)$

Note that $x^{2} + 1$ $x^2+1$ has no simpler linear factors with real coefficients since $x^{2} + 1 \geq 1 > 0$ $x^2+1 >= 1 > 0$ for all $x in RR$

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