Factor as much as possible

A-R-Q · **Joined:** Sat, 20 Nov 2010 21:17:02 UTC **Posts:** 432

Factor as much as possible
c)

z^4 - 2

Attempt:
<=> $(z^2)^2 - (2^{1/2})^2$
<=> $(z^2 - 2^{1/2})(z^2 + 2^{1/2})$

<=> $((z)^2 - (2^{1/4})^2)(z^2 + 2^{1/2})$
<=> $(z - 2^{1/4})(z + 2^{1/4})(z^2 + 2^{1/2})$

Can I go further than this, or is this enough?

Shadow · **Posted:** Thu, 24 Nov 2011 05:39:20 UTC

A-R-Q wrote:

Factor as much as possible
c)

z^4 - 2

Attempt:
<=> $(z^2)^2 - (2^{1/2})^2$
<=> $(z^2 - 2^{1/2})(z^2 + 2^{1/2})$

<=> $((z)^2 - (2^{1/4})^2)(z^2 + 2^{1/2})$
<=> $(z - 2^{1/4})(z + 2^{1/4})(z^2 + 2^{1/2})$

Can I go further than this, or is this enough?

It depends on what ring you're factoring over. If it's just the integers, you can't do anything, if you're willing to allow the square root of 2, you can go farther, if you're willing to allow complex numbers, you can go even farther.

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Factor as much as possible

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