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.
 A: Consider
$$p(x) = x^5-(3-a)x^4+(3-2a)x^3-ax^2+2ax-a$$ 
First, we note that if $x< 0$, each term is negative, hence there are no negative roots.  Also, $p(0) = -a < 0$.  Further,
$$p'(x) = 5x^4-4(3-a)x^3+3(3-2a)x^2-2ax+2a$$
So it is sufficient to show that $p'(x) > 0$ for $x > 0, \; a \in (0, \frac12)$.  
For this, note that by AM-GM inequality, $\frac12 ax^2+2a \ge 2ax$, so it is sufficient to show that:
$5x^4+\frac12(18-13a)x^2 > 4(3-a)x^3$.  By AM-GM we again have:
$$5x^4+\tfrac12(18-13a)x^2 \ge 2\sqrt{\frac{5(18-13a)}2}x^3$$
So it is enough to show $5(18-13a) > 8(3-a)^2 \iff 8a^2+17a < 18$, which is true for $a \in (0, \frac12)$.
A: 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?
