<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] $(x^3+x+1)'=3x^2+1>0$ for all real $x$.
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