Proving roots of a function cannot all be real

Question

The question is

Let $a$$,b$$,c$$,d$ be any four real number not all equal to zero. Prove that the roots of the polynomial $f(x)=x^6+ax^3+bx^2+cx+d$ cannot all be real.

I know that the first derivative of a functions gives its turning points and the second derivative give its concavity but how does it help to solve the question. Or if you have any other methods it would really help.Any hint would also be appreciated . Thanks in advance.

Answer 1

If $f$ has $6$ real roots (perhaps with multiplicity), then $f'(x)$ has $5$ real roots, $f''(x)$ has $4$ real roots, and so on. We have $$ \begin{align} f(x)&=x^6+ax^3+bx^2+cx+d\\ f'(x)&=6x^5+3ax^2+2bx+c\\ f''(x)&=30x^4+6ax+2b\\ f'''(x)&=120x^3+6a\\ \end{align} $$ Note that $x^m+p$ with $m>2$ has $m$ real roots if and only if $p=0$ (because it can have at most two distinct real roots, and a multiple root only if $p=0$). Using this, we see from the expression for $f'''(x)$ that $a=0$. Then by the same argument using $f''(x)$, we have $b=0$. After that using $f'(x)$, we obtain $c=0$, and finally using $f(x)$, we find $d=0$. But $a=b=c=d=0$ is excluded in the problem statement.

Answer 2

Descartes rule of signs is independent of the degree sequence, it only demands that the coefficients under consideration are degree-ordered when computing the sign variations. This means that the given polynomial has the same maximal amount of positive, negative and zero roots as $0=x^4+ax^3+bx^2+cx+d$.

Which means that among the roots of the original degree-6 polynomial there are at most $4$ real roots, at least $2$ roots have to be non-real complex.