How do you factor $8 x^{6} - 32 x^{5} + 4 x^{4}$ $8x^6-32x^5+4x^4$ ?

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Answer 1 · 2015-05-13T15:33:35

$y = 4 x^{4} . (2 x^{2} - 8 x + 1)$ $y = 4x^4.(2x^2 - 8x + 1)$

The trinomial in parentheses can't be factored because $D = b^{2} - 4 a c = 64 - 8 = 56$ $D = b^2 - 4ac = 64 - 8 = 56$ is not a perfect square.

How do you factor 8x^6-32x^5+4x^48x6−32x5+4x48x^6-32x^5+4x^4?