Palindromic Polynomial

Question

What are the steps to solve this and what is the goal?

$$ax^6 + bx^5 + cx^4 + dx^3 + cx^2 + bx + a$$

What do you mean by solve? – Peter Foreman Aug 21 '19 at 10:11 — Peter Foreman, Aug 21 '19 at 10:11
Should there be an $=0$ at the end? – projectilemotion Aug 21 '19 at 12:17 — projectilemotion, Aug 21 '19 at 12:17

score 1 · Answer 1 · answered Aug 21 '19 at 11:06

1

Hint: $x=0$ is not a solution, so we can write $$a(x^3+\frac{1}{x^3})+b(x^2+\frac{1}{x^2})+c(x+\frac{1}{x})+d=0$$ and substitute $$t=x+\frac{1}{x}$$

answered Aug 21 '19 at 11:06

score 0 · Answer 2 · answered Aug 21 '19 at 14:07

0

Given: $x^6+ax^5+bx^4+cx^3+bx^2+ax+1=0$

Assuming a root's inverse is not conjugate, then it must one of four related roots that satisfy the quartic

$$x^4+ax^3+bx^2+ax+1$$

If $a^2-4b+8\gt 0$, the quartic may be split into two palindromic quadratics.

answered Aug 21 '19 at 14:07

poetasis

2 Answers2