MHB Roots of polynomial equations ( Substitution )

AI Thread Summary
The discussion focuses on reducing the polynomial equation u^4 + 5u^3 + 6u^2 + 5u + 1 = 0 to the simpler form v^2 + 5v + 4 = 0 using the substitution v = u + 1/u. Participants suggest dividing the original equation by u^2 to facilitate the transformation. It is noted that substituting v into the simpler equation yields the original polynomial, confirming the validity of the approach. Additionally, similar techniques are proposed for solving more complex polynomial equations. The thread emphasizes the effectiveness of substitution in polynomial reduction.
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 $$x^6-9x^5+29x^4-42x^3+29x^2-9x+1=0$$.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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