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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.

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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.

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    $\begingroup$ I am trying to learn. Using the derivative is useful as a test for multiple roots, and in some cases to find a multiple root. In this question, we were not told that there are equal roots. Why did you use the derivative as the base for this answer? Thanks. $\endgroup$ – NoChance Oct 2 '19 at 3:38
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    $\begingroup$ @NoChance: "If $f(x)$ has $m$ real roots, then $f'(x)$ has $m-1$ real roots" is basically Rolle's theorem plus a little bookkeeping. $\endgroup$ – Micah Oct 2 '19 at 4:22
  • $\begingroup$ @Micah, thank you for your comment but I am not sure how Rolle's theorem is to do with this. What I don't understand is why did the answer assume that f(x) has repeated roots. $\endgroup$ – NoChance Oct 2 '19 at 6:18
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    $\begingroup$ The answer doesn't assume that there are repeated roots; it accommodates the possiblity that there are repeated roots. $\endgroup$ – Greg Martin Oct 2 '19 at 6:25
  • $\begingroup$ @GregMartin, thanks for your comment. $\endgroup$ – NoChance Oct 2 '19 at 10:13
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Suppose all the roots of $f(x)$ are real.

By Vieta's formula,

$$\text{Sum of the roots}=-\sum \alpha = 0.$$

Squaring, $$\sum \alpha^2 + 2 \sum \alpha \beta = \sum\alpha^2 + 0 = 0.$$

This is possible only if all the roots are equal to $0$, which is false since $a, b, c, d$ are not all $0$ .

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    $\begingroup$ $-\sum\alpha$ is the coefficient of $x^{n-1}$, and $\sum\alpha\beta$ is the coefficient of $x^{n-2}$. $\endgroup$ – Hagen von Eitzen Oct 1 '19 at 15:36
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    $\begingroup$ Fixed thank you $\endgroup$ – user600016 Oct 1 '19 at 16:08
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    $\begingroup$ -1, this answer is too confusing for a non-expert. If it is improved with a prelude that explains the notation and/or theorems being used I'll retract my downvote. $\endgroup$ – goblin Oct 2 '19 at 2:41
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    $\begingroup$ So please add that information and a link in the answer for those that do not know what you are thinking. $\endgroup$ – StephenG Oct 2 '19 at 2:58
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    $\begingroup$ Im more curious about the "squaring" comment. Squaring what? Where does beta come from? No explanatory content whatsoever. $\endgroup$ – SquishyRhode Oct 2 '19 at 3:18
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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.

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