Newton Map in Dynamical Systems Let $$p(x)=(x^2-1)(x^2-4)=x^4-5x^2+4$$. Let $$N(x)=Np(x)$$. Notice that N(x) goes to +-infinity at $$a1=-\sqrt{2.5}$$ $$a2=0$$ $$a3=\sqrt{2.5}$$
a. Sketch the graph of N(x).
So the Newton map would be
$$Np(x)=x-\frac{p(x)}{p'(x)}$$
$$Np(x)=x-\frac{x^4-5x^2+4}{4x^3-10x}$$
$$Np(x)=\frac{x(4x^3-10x)}{4x^3-10x}-\frac{x^4-5x^2+4}{4x^3-10x}$$
$$Np(x)=\frac{4x^4-10x^2-x^4-5x^2+4}{4x^3-10x}$$
$$Np(x)=\frac{3x^4-15x^2+4}{4x^3-10x}$$
But I don't really know how to simplify it further/if this is correct/how to graph it?
 A: I think you made a little mistake by not distributing the minus sign through when going from your third line to your fourth line.  Thus, I get
$$N(x) = \frac{3 x^4-5 x^2-4}{4 x^3-10 x}.$$
I don't think that it needs to be simplified further; there's countless ways you could rewrite it, I suppose.
As far as the graph goes, it clearly has vertical asymptotes when the denominator equals zero at $x=0,\pm\sqrt{5/2}$.  If you take the derivative of $N(x)$, you should get
$$N'(x)=\frac{6 x^6-35 x^4+49 x^2-20}{2 x^2 \left(5-2x^2\right)^2}.$$
Now, the denominator of $N'$ is always non-negative and the numerator is negative on $-2<x<2$. Thus, $N$ is decreasing between its vertical asymptotes and asymptotic to $3x/4$ for large $x$.  I guess you have to get something like so:

The yellow line is the line $y=x$ and we see intersections at the roots of the original polynomial, as expected. Note that $N(x)$ crosses $y=x$ with slope zero. This is exactly why those points are super-attracting fixed points of $N$, explaining why Newton's method converges so rapidly.
Just for the fun of it, and because Amzoti had mentioned that you might be interested in the complex plane, here's an illustration of the basins of attraction of the roots of your polynomial in the complex plane.

