Author Topic: Homework Help :) (Read 3933 times) Tweet Share

rainbows. · « **Reply #15 on:** March 21, 2010, 10:05:06 pm »

OHHH someone help me factorise Z^6 - 64 please

kamil9876 · « **Reply #16 on:** March 21, 2010, 10:15:14 pm »

solve $z^6-64=0$ (Polar form works really sweet for this one)

You should get 6 roots $z_1,z_2...z_6$

Therefore $z^6-64=(z-z_1)(z-z_2)...(z-z_6)$

Mao · « **Reply #17 on:** March 22, 2010, 02:43:21 am »

moekamo · « **Reply #18 on:** March 22, 2010, 02:52:11 am »

Quote from: Mao on March 22, 2010, 02:43:21 am

$z^6-64 = (z^3 + 8)(z^3-8) = (z+2)(z^2-2z+4)(z-2)(z^2+2z+4) = (z-2)(z+2)(z-1+\sqrt{3}i)(z-1-\sqrt{3}i)(z+1+\sqrt{3}i)(z-1-\sqrt{3}i)$

you have two $(z-1-\sqrt{3}i)$ factors in here... shouldnt one of them be $(z+1-\sqrt{3}i)$

Mao · « **Reply #19 on:** March 22, 2010, 03:06:40 am »

Quote from: moekamo on March 22, 2010, 02:52:11 am

Quote from: Mao on March 22, 2010, 02:43:21 am
$z^6-64 = (z^3 + 8)(z^3-8) = (z+2)(z^2-2z+4)(z-2)(z^2+2z+4) = (z-2)(z+2)(z-1+\sqrt{3}i)(z-1-\sqrt{3}i)(z+1+\sqrt{3}i)(z-1-\sqrt{3}i)$

you have two $(z-1-\sqrt{3}i)$ factors in here... shouldnt one of them be $(z+1-\sqrt{3}i)$

yes, that'll teach me about copy/pasting tex codes.

the.watchman · « **Reply #20 on:** March 22, 2010, 06:48:49 am »

Quote from: Mao on March 22, 2010, 03:06:40 am

Quote from: moekamo on March 22, 2010, 02:52:11 am
Quote from: Mao on March 22, 2010, 02:43:21 am
$z^6-64 = (z^3 + 8)(z^3-8) = (z+2)(z^2-2z+4)(z-2)(z^2+2z+4) = (z-2)(z+2)(z-1+\sqrt{3}i)(z-1-\sqrt{3}i)(z+1+\sqrt{3}i)(z-1-\sqrt{3}i)$

you have two $(z-1-\sqrt{3}i)$ factors in here... shouldnt one of them be $(z+1-\sqrt{3}i)$

yes, that'll teach me about copy/pasting tex codes.

LOL i do that all the time...

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Author Topic: Homework Help :) (Read 3933 times) Tweet Share

rainbows.

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