<s>[INST] 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)
 [/INST] 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.
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