anonymous
  • anonymous
Solve x^4+x^3+x^2+x+1 = 0 by hand
Meta-math
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Mr.Math
  • Mr.Math
You're studying complex numbers I assume. I will give you a hint: \[x^4+x^3+x^2+x+1=\frac{x^5-1}{x-1}.\] Solve it now by first finding the roots of unity of \(x^5=1\).
anonymous
  • anonymous
Why is x^4+x^3+x^2+x+1 = (x^5-1)/(x-1)
Mr.Math
  • Mr.Math
This come from the factorization \(x^5-1=(x-1)(x^4+x^3+x^2+x+1)\). If you don't already know that. Then you can easily notice that \(1\) is a root of \(x^5-1\), and then you can use synthetic division to get the above factorization.

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Mr.Math
  • Mr.Math
comes*
anonymous
  • anonymous
o right
anonymous
  • anonymous
but does that give you all of the roots?
Mr.Math
  • Mr.Math
\(x^5-1\) has all the roots of your problem and one more root which is \(x=1\), so you have to exclude it.
Mr.Math
  • Mr.Math
You know how to solve \(x^5=1\)?
anonymous
  • anonymous
1 and 4 complex solutions
Mr.Math
  • Mr.Math
Exactly. The four complex solutions are what you're looking for, as you can see from the factorization of \(x^5-1\).
anonymous
  • anonymous
im not really sure about how to solve x^5=1
Mr.Math
  • Mr.Math
Have a look at this http://www.youtube.com/watch?v=9Z1uNz__2xA
anonymous
  • anonymous
o ty
anonymous
  • anonymous
Or First divide the entire equation by x^2 \[x^2+x+1+\frac{1}{x}+\frac{1}{x^2}=0 \implies x^2+\frac{1}{x^2}+x+\frac{1}{x}+1=0\] Let u = x+1/x \[u^2 = x^2+2+\frac{1}{x^2}\] The original equation now becomes: \[u^2-2+u+1 = 0 \implies u^2+u-1=0\] Use quadratic equation then substitute u = x+1/x back and then solve the quadratic again.

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