$2x^4+x^3+mx^2+x+2=0$ has only real roots $2x^4+x^3+mx^2+x+2=0$, $m\in \Bbb R$. For which interval that defines m does the polynomial have ONLY real roots.
Tried some Horner rule, Vieta's formulas and Sturm's theorem(although I'm not familiar with the last one) but I didn't seem to get anywhere with none of these..
Any hints on how I should approach it? Thank you
 A: Rewrite the equation as
$$
2x^2+\frac{2}{x^2}+x+\frac{1}{x}+m=0
$$
which becomes
$$
2\left(x+\frac{1}{x}\right)^{\!2}+\left(x+\frac{1}{x}\right)+m-4=0
$$
This equation in $t=x+\frac{1}{x}$ should have real roots, so
$$
33-8m\ge0
$$
If $d=\sqrt{33-8m}$, then the two equations
$$
x+\frac{1}{x}=\frac{-1+d}{4}
$$
and
$$
x+\frac{1}{x}=\frac{-1-d}{4}
$$
should have real roots too.
A: Since $f(0)=2\gt 0$ and the degree of $f$ is $4$ with coefficient of $x^4$ positive it is enough to consider $f(x)$ has a double root and this root be positive.
Making $$f(x)=(x-a)^2(2x^2+bx+c)$$ this gives $$f(x)==2x^4+(b-4a)x^3+(c-2ab+2a^2)x^2+(a^2b-2ac)x+a^2c$$ from which the system in integers
$$\begin{cases}b-4a=1\\a^2b-2ac=1\\a^2c=2\end{cases}$$ whose solution is$$(a,b,c)=(1,5,2)$$ Consequently $$m=2-10+2=-6$$ It is easy now to verify that $\color{red} {m\le-6}$ works.
In fact, the double root of $$f(x)=2x^4+x^3-6x^2+x+2$$ is $1\gt0$ the other two roots being negative. 
A: necessary condition
It will have 4 real roots.
By Rolle's Theorem, the derivative will have 3 roots and the second derivative should have two roots.
$$f''(x)=24x^2+6x+2m $$
the discriminant must be positive.
$$\Delta=9-48m>0$$
$$\iff m <\frac {3}{16} $$
A: Let $x+\frac{1}{x}=u$.
Hence, $|u|\geq2$ and we have $2u^2+u+m-4=0.$
Let $f(u)=2u^2+u+m-4$.
For existing four real roots of the given equation we need that the equation $f(u)=0$
will be have two real roots $u_1$ and $u_2$.
Let $u_1\leq u_2$.
Since $-2<-\frac{1}{4}<2$ (see on the vertex of the parabola), we need $u_1\leq-2$ and $u_2\geq2$,  
which gives  $f(2)\leq0$ and $f(-2)\leq0$, which gives $m\leq-6$.
Done!
