cubic equations which have exactly one real root Question is to check :
For any real number $c$, the polynomial $x^3+x+c$ has exactly one real root .
the way in which i have proceeded is :
let $a$ be one real root for $x^3+x+c$ i.e., we have $a^3+a+c=0$
i have seen that $(x-a)(x^2+ax+(a^2+1))=x^3+x-a^3-a$
But, $a^3+a+c=0$. So, $-a^3-a=c$.
so, $(x-a)(x^2+ax+(a^2+1))=x^3+x-a^3-a=x^3+x+c$
i.e, $(x-a)(x^2+ax+(a^2+1))=x^3+x+c$
Now, determinant for quadratic $x^2+ax+(a^2+1)$ is $a^2-4(a^2+1)=-3a^2-4 <0$ for any real number $a$
Thus, quadratic has no real root and so is the cubic $x^3+x+c$
I would like to know if this justification is sufficient and if this can be generalized.
I mean, can i say $x^3+ax^2+bx+c$ for $a,b,c\in \mathbb{Z}$ have exactly one real root. (Conditions apply)
can this be generalized to any odd degree polynomial (at least for some special cases)
 A: You essentially split off the linear factor belnging to a real root and showed per determinant that the remaining quadratic has no real root.
This is fine but does not readily generalize to higher degrees.
Instead, it is probably easier to show that $f$ is injective: Assume $a,b$ are two real roots, i.e. $f(a)=f(b)=0$.
Then by Rolle, there exists $\xi$ between $a$ and $b$ (including the case $a=\xi=b$ if you also want to show that no multiple root - $a=b$ - exists) with $f'(\xi)=0$. But for $f(x)=x^3+x$ we have $f'(x)=3x^2+1\ge1$ for all $x$.
A: $(x^3+x+1)'=3x^2+1>0$ for all real $x$.
A: It is a bit unclear what you want to generalise. $x^3+x+c$ is a strictly increasing function of $x$ - a strictly increasing (or decreasing) polynomial of odd order with real coefficients will always have a single real root.
Or you may be saying that the polynomial can be written as $p(x)=(x-a)q(x)$ where $q(x)$ is a polynomial of even degree which has no real roots. Any odd degree polynomial with a single simple zero (ie not a multiple zero) can be written in this way. However this form of factorisation does not imply that $p(x)$ is strictly increasing, simply that it has just one real zero.
For example take $p(x)=x\left((x-3)^2+1\right)$ with $a=0$
We have $p(0)=0, P(1)=5, p(2)=4, p(3)=3, p(4)=8$, so $p(x)$ is not increasing, while $q(x)$, which is a sum of squares, one of which is non-zero, has no real roots.
