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 $g= X^5+9X^3+3X-15$. Where $E=\mathbb{Q}$($\alpha$).
I have determined $\mathbb{Q}$($\alpha$) ={ $\sum^4_0 a_i$ $\alpha^i$, $i$ $\in$ $\mathbb{Q}$ }. I.e. $a_0$ +$a_1\alpha^1$...+$a_4\alpha^4.$ 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?
Thankyou in advance.
Consider the function $g: \mathbb{R} \to \mathbb{R}$ defined by $g(x) = x^5 + 9x^3 + 3x - 15$.
Observe $g'(x) = 5x^4 + 27x^2 + 3$, which is always positive: an even power is greater than or equal to zero, and you are summing two even powers along with the number $3$. So, in particular, you have that $g'(x) \geq 3$ for all $x \in \mathbb{R}$, which means that $g$ is always increasing.
But if $g$ is monotonically increasing, then (noting it takes on both negative and positive values) it can only have one $x$-intercept, which corresponds to only one real root. Let $\alpha \in \mathbb{R}$ be this root. Then the other four roots are all imaginary, which means, in particular, none of the four imaginary roots is contained in $\mathbb{Q}(\alpha) \subset \mathbb{R}$.