Four real roots of $x^4+2x^3+mx^2+2x+1=0$ iff $m$ is... The Equation $x^4+2x^3+mx^2+2x+1=0$ has $4$ different real roots iff:
a) $m<3$;
b) $m<2$;
c) $m<-6$;
d) $1<m<3$;
e) $-6<m<2$  

Notice that $x=0$ is not going to be a solution. Then notice the coefficients are symmetric. This means that if we divide by $x^{4/2}$ we get $$(x^2+1/x^2)+2(x+1/x)+m=0$$
which can be written as
$$(x+1/x)^2+(x+1/x)+m-2=0.$$
Putting $y:=x+1/x$ we get $$y^2+y+m-2=0$$
which has two different real solutions for $1-4(m-2)>0$, i.e. $9>4m$. For these we have $x+1/x=y=\frac{-1\pm\sqrt{1-4(m-2)}}{2}$.
These gives us two polynomials of degree two:
$$x^2-\frac{-1\pm\sqrt{1-4(m-2)}}{2}x+1.$$
To get two different real solutions out of these we need again positive discriminant $\left(\frac{-1\pm\sqrt{1-4(m-2)}}{2}\right)^2-4>0$.
