Proof polynomial has only one real root. I need to prove that this polynomial equation:
$$x^5-(3-a)x^4+(3-2a)x^3-ax^2+2ax-a=0\quad\text{ for }\quad a\in(0,\frac{1}{2}).$$
has only one root. That it has one real root is obvious because it is of odd degree. But Descartes rules here fails to bound the number of roots to one.

I try to factorize your formula :
$x^5-(3-a)x^4+(3-2a)x^3-ax^2+2ax-a=0\quad\text{ for }\quad a\in(0,\frac{1}{2})$
$x^5-(3-a)x^4+(3-2a)x^3-ax^2+2ax-a=x^5+(3-a)(x^3-x^4)-ax^{3}-ax^{2}+2ax-a$
$=x^5+(1-a)(x^3-x^4)-ax^{3}-ax^{2}+2ax-a+2x^{3}-2x^{4}$
$=x^5-1+1+(1-a)(x^3-x^4)-a(x^{3}-x^{2})+2ax+2ax^{2}-a+2x^{3}-2x^{4}$
$=x^5-1+(1-a)(x^3-x^4)-a(x^{3}-x^{2})+2a(x-1)+2x^{3}(1-x)+1+a+2ax^{2}$
$=(x-1)(x^{4}+x^{2}+x+1+ax^{3}-ax^{2}+2a+2x^{2})+1+a+2ax^{2}$$=x(x^{4}+x^{2}+x+1+ax^{3}-ax^{2}+2a+2x^{2})+1+a-(x^{4}+x^{2}+x+1+ax^{3}-ax^{2}+2a+2x^{2}-2ax^{2})$
$=x(x^{4}+x^{2}+x+1+ax^{3}-ax^{2}+2a+2x^{2})-(x^{4}+x^{2}+x+ax^{3}-ax^{2}+a+2x^{2}-2ax^{2})$
$\Longrightarrow{x(x^{4}+(a-1)x^{3}+(3-2a)x^{2}+(3a-2)x+2a)=a}$
$\Longrightarrow{f(x)=x(x^{4}+(a-1)x^{3}+(3-2a)x^{2}+(3a-2)x+2a)}$
$\Longrightarrow{f^{'}(x)=0}$
$\Longrightarrow{-10x^{4}=2-4a>0}$
a contradiction shows that $x=0$ is its unique root!
Is it helpful?
