jmc

We see that $x=0$ cannot be a root of the polynomial. Dividing both sides by $x^2,$ we get $$x^2+2px+1+\frac{2p}{x}+\frac{1}{x^2}=0.$$Let $y=x+\frac{1}{x}.$ Then $$y^2=x^2+2+\frac{1}{x^2},$$so $$y^2-2+2py+1=0,$$or $y^2+2py-1=0.$ Hence, $$p=\frac{1-y^2}{2y}.$$If $x$ is negative, then by AM-GM, $$y=x+\frac{1}{x}=-\left(-x+\frac{1}{-x}\right)\le -2\sqrt{(-x)\cdot \frac{1}{-x}}=-2.$$Then $$\frac{1-y^2}{2y}-\frac{3}{4}=\frac{-2y^2-3y+2}{4y}=-\frac{(y+2)(2y-1)}{4y}\ge 0.$$Therefore, $$p=\frac{1-y^2}{2y}\ge \frac{3}{4}.$$If $y=-2,$ then $x+\frac{1}{x}=-2.$ Then $x^2+2x+1=(x+1{)}^2=0,$ so the only negative root is $-1,$ and the condition in the problem is not met. Therefore, $y<-2,$ and $p>\frac{3}{4}.$

On the other hand, assume $p>\frac{3}{4}.$ Then by the quadratic formula applied to $y^2+2py-1=0,$ $$y=\frac{-2p\pm \sqrt{4p^2+4}}{2}=-p\pm \sqrt{p^2+1}.$$Since $p>\frac{3}{4},$ $$\begin{array}{rl}-p-\sqrt{p^2+1}& =-(p+\sqrt{p^2+1})\\ & <-\left(\frac{3}{4}+\sqrt{{\left(\frac{3}{4}\right)}^2+1}\right)\\ & =-2.\end{array}$$In other words, one of the possible values of $y$ is less than $-2.$

Then from $y=x+\frac{1}{x},$ $$x^2-yx+1=0.$$By the quadratic formula, $$x=\frac{y\pm \sqrt{y^2-4}}{2}.$$For the value of $y$ that is less than $-2,$ both roots are real. Furthermore, their product is 1, so they are both positive or both negative. The sum of the roots is $y,$ which is negative, so both roots are negative, and since $y^2-4\ne 0,$ they are distinct.

Therefore, the value of $p$ that works are $$p\in \boxed{\left(\frac{3}{4},\infty \right)}.$$