Compute the largest root of $x^4-x^3-5x^2+2x+6$ I want to calculate the largest root of $p(x)=x^4-x^3-5x^2+2x+6$. I note that $p(2) = -6$ and $p(3)=21$. So we must have a zero between two and three. Then I can go on calculating $p(\tfrac52)$ and see that the zero must lie in the interval $]2,\tfrac52[$. The answer is $\frac{1+\sqrt{13}}{2}$. But how can I derive that exact result?
Furthermore, to know that it is the largest root, I must find the other two roots, must not I? By sign shift I know one root between $1$ and $2$, but this method cannot give me the third root (the graph touches the $x$-axis). Maybe calculus? Using calculus we get a degree polynomial of degree which also is hard to find roots to. We get $p'(-1)=5$ and $p'(-2)=-22$ so we have a stationary point in $]-2,-1[$ and similarly we get one in $]1,2[$. Since it is two stationary points it must have one maximum and one minimum point. But how do I know which is which, I mean I cannot plug in a value in the second derivative. 
Can somebody help me?
Thanks in advance

You've : $p(x) = x^4 - x^3 - 5x^2 + 2x + 6 $ .
Rearrange the given polynomial as : $$ p(x) =(x^4 - 5x^2 + 6) - x(x^2 - 2) $$
For the first term, let us factor it separately. Say,  $x^2 = t$ . So, you've : 
$$\begin{align} f(x) =&\ x^4 - 5x^2 + 6\\ 
f(x) =&\ t^2 - 5t + 6 \end{align}$$ 
Roots of $f(x)$ are : $t = 3 \ \text{OR} \ 2 $ . Now, plugging-in $t = x^2$ back, we get: $x = \pm \sqrt{3},\pm \sqrt{2}$ .
So, we can write $f(x)$ as : $f(x) = (x-\sqrt{3})(x+\sqrt{3})(x-\sqrt{2})(x+\sqrt{2})$ 
And thus, $p(x)$ can be written as: 
$$\begin{align} p(x) =& (x-\sqrt{3})(x+\sqrt{3})(x-\sqrt{2})(x+\sqrt{2}) - x(x^2 -2) \\
=& (x^2 - 3)(x^2 -2 ) -x(x^2-2) \\
=& (x^2-2)(x^2 - x - 3) \end{align}$$
Therefore, you've : $p(x) = (x^2 - 2)(x^2 - x - 3)$ 
You can proceed from here, I guess. Hope it helps.
