How do you factor $8 x^{4} - 128$ $8x^4-128$ ?

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How do you factor $8 x^{4} - 128$ $8x^4-128$ ?

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Answer 1 · 2015-12-26T15:54:30

It is

$8 x^{4} - 128 = 8 \cdot (x^{4} - 16) = 8 \cdot (x^{4} - 2^{4}) = 8 \cdot (x^{2} - 4) \cdot (x^{2} + 4) = 8 \cdot (x - 2) \cdot (x + 2) \cdot (x^{2} + 4)$ $8x^4-128=8*(x^4-16)=8*(x^4-2^4)=8*(x^2-4)*(x^2+4)=8*(x-2)*(x+2)*(x^2+4)$

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How do you factor $8 x^{4} - 128$ $8x^4-128$ ?

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