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If x^6+1=(x^3+nx+1)(x^3-nx+1), n=?

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If x^6+1=(x^3+nx+1)(x^3-nx+1), n=?

by Max@Math Revolution » Thu Sep 29, 2016 5:24 am
If x^6+1=(x^3+nx+1)(x^3-nx+1), n=?
A. ±1/√2 B. ±1 C. ±√2 D. ±√3 E. ±2
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by regor60 » Thu Sep 29, 2016 8:42 am
Max@Math Revolution wrote:If x^6+1=(x^3+nx+1)(x^3-nx+1), n=?
A. ±1/√2 B. ±1 C. ±√2 D. ±√3 E. ±2
n=(2x)^1/2.

None of your answer choices work
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by Max@Math Revolution » Sat Oct 01, 2016 1:06 am
==>If you substitute x=1, from 1^6+1=(1+n+1)(1-n+1), you get 2=(2+n)(2-n)=4-n^2. Naturally, you get n^2=4-2=2, and n=±√2. Therefore the answer is C.

Answer: C
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by regor60 » Sun Oct 02, 2016 12:03 pm
Max@Math Revolution wrote:==>If you substitute x=1, from 1^6+1=(1+n+1)(1-n+1), you get 2=(2+n)(2-n)=4-n^2. Naturally, you get n^2=4-2=2, and n=±√2. Therefore the answer is C.

Answer: C
Maybe I'm slow. X is a variable, so the solution has to work for all X. So if you chose X=2,then n would equal 2.

What's the basis for assigning 1 to X?
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by Max@Math Revolution » Thu Oct 13, 2016 6:09 pm
Hi,

The question is supposed to be "x^4+1=(x^2+nx+1)(x^2-nx+1), n=?".
Sorry about the mistake.
Thank you.
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