I am looking for a detailed step by step answer if possible as I have an exam tomorrow a past paper questions is given by:
Let a real number $\alpha$ be a root of the polynomial $g= X^{193}+49X^2+14X+14$. Where $E=\mathbb{Q}(\alpha)$.
I have determined $\mathbb{Q}$($\alpha) =\bigl\{ \sum^4_0 a_i\alpha^i\mid i \in \mathbb{Q}\bigr\}$. I.e., $a_0+a_1\alpha^1+\cdots+a_{192}\alpha^{192}$. I have also shown $g$ is irreducible in $\mathbb{Q}$, However I am unclear how to show how the only root of $g$ in $E$ is $\alpha$. My lecturer hinted at showing the polynomial only crosses the $x$ axis once may you expand?
Thank you in advance.
If we believe for a moment that $g$ has only one real root, then the fact that it has only one root in $\Bbb Q(\alpha)$ is easy: $\Bbb Q(\alpha)$ is a subfield of $\Bbb R$, so any root in $\Bbb Q(\alpha)$ must be real, and therefore equal to$~\alpha$.
Now to show that indeed $g$ has only one real root, consider first the values on the interval $[-1,1]$, where $X^{193}$ evaluates with absolute value at most$~1$. Certainly $g(0)=14$ is positive, so if one can show that $49x^2+14x+13>0$ for all $x\in[-1,1]$, then $g$ will have no roots in that interval (either). But that is a quadratic with discriminant $14^2-4\times49\times13<0$, so indeed it fails to have any real roots at all. So we know all roots of $g$ must be outside $[-1,1]$. One clearly has $g(x)>0$ for all $x>1$, so we only need to consider roots $x<-1$. Now $g'(-1)=193-98+14>0$ and $g''(x)=193\times192x^{191}+49\times48<0$ for all $x\leq-1$, from which it follows that $g'(x)>0$ for all $x\leq-1$, so below $-1$, the function $x\to g(x)$ is strictly increasing and can have only one root (which of course it does have).