Roots of polynomial equations ( Substitution )

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SUMMARY

The discussion focuses on reducing the polynomial equation u4 + 5u3 + 6u2 + 5u + 1 = 0 to the simpler form v2 + 5v + 4 = 0 using the substitution v = u + 1/u. Participants confirm that dividing through by u2 is a valid step, leading to the desired transformation. Additionally, the discussion introduces related polynomial equations such as x6 - 6x5 + 14x4 - 18x3 + 14x2 - 6x + 1 = 0, demonstrating the application of the same substitution technique.

PREREQUISITES
  • Understanding polynomial equations and their roots
  • Familiarity with algebraic substitutions
  • Knowledge of manipulating equations
  • Basic skills in solving quadratic equations
NEXT STEPS
  • Study the method of polynomial substitution in depth
  • Learn how to apply the quadratic formula to solve v2 + 5v + 4 = 0
  • Explore advanced polynomial equations and their transformations
  • Investigate the implications of dividing by variables in polynomial equations
USEFUL FOR

Mathematicians, algebra students, and educators looking to deepen their understanding of polynomial transformations and substitutions in algebraic equations.

Erfan1
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How do I reduce u^4 + 5u^3 + 6u^2 + 5u + 1 = 0 to v^2 + 5v + 4 = 0 by using v = u + 1/u ?
 
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Erfan said:
How do I reduce u^4 + 5u^3 + 6u^2 + 5u + 1 = 0 to v^2 + 5v + 4 = 0 by using v = u + 1/u ?

Since $$u \ne 0$$ you may divide through by $$u^2$$.

What do you notice now?
 
Erfan said:
How do I reduce u^4 + 5u^3 + 6u^2 + 5u + 1 = 0 to v^2 + 5v + 4 = 0 by using v = u + 1/u ?
You can do as M R suggested or you can put $v=u+1/u$ in $v^2+5v+4=0$. You should get $u^4 + 5u^3 + 6u^2 + 5u + 1 = 0$.
 
You can use the same idea to solve $$x^6-6x^5+14x^4-18x^3+14x^2-6x+1=0$$, which I made specially for you. :)
 
And a slightly nicer one $$x^6-9x^5+29x^4-42x^3+29x^2-9x+1=0$$.
 

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