How can I convince students that $p(x)=0$ is a symmetric equation if they ask me, where
$p(x)$ is polynomial of degree $n$ with reals coefficients. For example :
$A(x)=2x^4-9x^3+8x^2-9x+2=0 $ is symmetric equation can be solved in $\mathbb{R}$ by taking
$z=x+\frac{1}{x}$ and $x \neq 0$ .
Note : $z$ and $x$ are reals numbers .
I would be interested in any replies or any comments. Thank you.
They do not say anything about Berber or French, however: http://en.wikipedia.org/wiki/Palindrome#Palindromes_in_Arabic_language
Yes, if you divide a palindromic polynomial through by the middle power of the variable, you get your substitution:
$$ 2 x^2 - 9 x + 8 - \frac{9}{x} + \frac{2}{x^2}. $$ Next use $$ \left( x + \frac{1}{x} \right)^2 = x^2 + 2 + \frac{1}{x^2} $$ and there is no difficulty writing the original problem in terms of $$ w = x + \frac{1}{x} $$
Less sure what happens in odd degree; $x^3 - 7 x^2 - 7 x + 1.$ Wait: guaranteed that $x=-1$ is a root. $(x+1)(x^2 - 8x + 1).$ So, if we start with $$ x^5 -37 x^4 + 23 x^3 + 23 x^2 - 37 x + 1 = (x+1)(x^4 - 38 x^3 + 61 x^2 - 38 x + 1). $$ LATHER, RINSE, REPEAT.