We have the polynomial $f(x)=x^3+6x-14 \in \mathbb{Q}[x]$. We have that $f(x)$ has exactly one positive real root $a$. That means that $f(x)$ can be written as followed:
$$f(x)=(x-a)(x^2+px+q)$$
Where do $p,q$ belong to?? Are they in $\mathbb{Q}, \mathbb{R}$ ??
$f(x)$ has exactly one real root because it is a strictly increasing function (no derivative required here: it's the sum of 3 increasing functions, one at least being strictly so).
By the intermediate Value Theorem, this root is between $1$ and $2 $: $f(1)=-7$, $f(2)=6$.
For a rational root $x=\dfrac n d$, we have that $n$ is a divisor of the constant term ($14$) and $d$ is a divisor of the dominant coefficient ($1$). Hence, if $x$ is rational, it is one of $\{\pm 1, \pm 2, \pm 7, \pm14\}$.
The only value between $1$ and $2$ is $2$, and it is not a root. Hence $x$ is irrational.
Note. Since there's only one real root, you can solve the equation with Cardano's method. One finds $$x=\sqrt[3]{7-\sqrt{57}}+\sqrt[3]{7+\sqrt{57}}\approx 1.622.$$
$$f(x)=(x-\xi)(x^2+ax+b), a, b \in \mathbb{Z}_2(\xi)$$
– Mary Star Jan 01 '15 at 22:11field because $x^3+x+1$ is irreducible over $\mathbf F_2$, and the root is simply the congruence class of $t$ modulo $t^3+t+1$ (cf. the construction of $\mathbf C$ as $\mathbf R[x]/(x^2+1)$). Now in the polynomial ring $\mathbf F_2[x]$, $x^3+x+1$ is divisible par $x-\xi$ since $\xi$ is a root of this polynomial.
– Bernard
Jan 01 '15 at 22:23
Though your question is a bit more specific, it is worth mentioning some general results about persistence of divisibility that often prove useful in such contexts.
A characteristic property of univariate polynomial rings over a field is that they are the only nontrivial rings with division-with-remainder, such that the quotient and remainder are unique.
Using said uniqueness it is very simple to prove that extending the coefficient field doesn't change divisibility results. So, e.g. if $\,f,g\in \Bbb Q[x],\,$ and $\,f\,$ divides $\,g\,$ in $\Bbb R[x]\,$ then $\,f\,$ divides $\, g\,$ in $\,\Bbb Q[x],\,$ i.e. the quotient $\,g/f\,$ lies in $\Bbb Q[x],\,$ i.e. its coefficients remain rational.
Another way to deduce such persistence of quotient (and remainder) is to start with the observation that the operations of the division algorithm are all rational over the coefficient field, so remain there. The advantage of the above view is that it sets it in the general context of deducing equalities from uniqueness theorems - a widely applicable method.
See said answer for further discussion and literature references.
$$f(x)=(x-\xi)(x^2+ax+b), a, b \in \mathbb{Z}_2(\xi)$$
They are in $\mathbb{Z}_2$ because we are in this field... But why is it $\mathbb{Z}_2(\xi)$ ??
– Mary Star Jan 01 '15 at 22:25The rational root theorem says that any rational root must be an integer divisor of $14$. But $f(x)=x^3+6x-14$ is strictly increasing, and we see that $f(1)<0$ and $f(2)>0$, so $f$ has not rational root.
That means $p,q$ are not rational, but are real.
$$f(x)=(x-\xi)(x^2+ax+b), a, b \in \mathbb{Z}_2(\xi)$$
– Mary Star Jan 01 '15 at 22:11We know $p,q \in \mathbb{R}$, and if $a \in \mathbb{Q}$, then $p,q \in \mathbb{Q}$ as well.
$$f(x)=(x-\xi)(x^2+ax+b), a, b \in \mathbb{Z}_2(\xi)$$
Do we have to apply again the long division algorithm??
– Mary Star Jan 01 '15 at 22:20$$f(x)=x^3+(p-a)x^2+(q-ap)x-aq$$
Then, $p=a$, because $f$ has no term on $x^2$. Now, $$f(x)=x^3+(q-a^2)x-aq$$ so $q-a^2=6$, $aq=14$.
Therefore, $a$ is rational if and only if $p$ is rational. This is obvious since they are the same number. Also, $q$ is rational if and only if $a$ is rational, because their product is a non zero rational number (namely, $14$).
$$f(x)=(x-\xi)(x^2+ax+b), a, b \in \mathbb{Z}_2(\xi)$$
Does it stand that $a, b \in \mathbb{Z}_2(\xi)$ because of the following??
$$f(x)=x^3+x^2(a-\xi)+x(b-a\xi)x-b \xi \Rightarrow a=\xi, b-a\xi=1 \Rightarrow b=1+\xi^2, -b \xi=1$$
So, $a$ and $b$ are expressions of $\xi$...
– Mary Star Jan 01 '15 at 22:17Since the root $a$ is described as a real root, then the factorisation is being described within $\mathbb R$ and $p, q$ should be assumed to be real numbers.
A related question is whether the original polynomial can be factorised within $\mathbb Q[x]$ - such a factorisation remains valid in $\mathbb R[x]$. This can only be the case if there is a rational root, and in this particular case any rational root must be an integer, and if there is a rational factorisation, there will be an integer factorisation (by a theorem of Gauss).
$$f(x)=(x-\xi)(x^2+ax+b), a, b \in \mathbb{Z}_2(\xi)$$
– Mary Star Jan 01 '15 at 22:22