[True/False]The polynomial $x^4+7x^3−13x^2+11x$ has exactly one real root.

I want to solve it without drawing the graph. Here is my idea. Note that $f(1)=1+7-13+11=6>0$ and $f(-1)=1-7-13-11=-30<0$

So we have at least one real root. Now since degree is $4$ we have $4$ roots but rest three can not be complex as they occur in pairs, so we must have another real root.

So the statement is False

Am I right?

Thanks for reading and all the help.

  • $\begingroup$ How do you know the root is not a double root? $\endgroup$ – Théophile Apr 29 '19 at 15:26
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    $\begingroup$ There are at least two. $0$ and one real root of $x^3+7x^2-13x+11$. $\endgroup$ – Saucy O'Path Apr 29 '19 at 15:26
  • $\begingroup$ You are right, but the root $x=0$ is even more obvious. $\endgroup$ – Yves Daoust Apr 29 '19 at 15:29

Yes, your result is correct (how to improve the actual argumentation is sufficiently discussed within the comments of this answer). It can be done in another by noting that $$f(x)=x^4+7x^3-12x^2+11x=x\cdot(x^3+7x^2-12x+11)=x\cdot g(x)$$ $f(x)$ has one real root at $x=0$. Now, $g(x)$ is of degree $3$ and every polynomial of degree $3$ has at least one real root by the Intermediate Value Theorem since $\lim\limits_{x\to-\infty}g(x)=-\infty$ and $\lim\limits_{x\to\infty}g(x)=\infty$.

Thus, $f(x)$ has at least two real roots.

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  • $\begingroup$ The answer that the OP reached is correct, but I feel that saying they are right is not entirely right, as the reasoning is very flawed. $\endgroup$ – Arthur Apr 29 '19 at 15:28
  • $\begingroup$ @Arthur Where could my approach possibly go wrong? I think my approach does not cover the fact the roots can have multiplicity 2 but even if that is the case, we would still have two real roots. right? It would be great of you, if you can clarify this last doubt of mine $\endgroup$ – StammeringMathematician Apr 29 '19 at 15:32
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    $\begingroup$ @StammeringMathematician Of course! Edited. $\endgroup$ – mrtaurho Apr 29 '19 at 15:34
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    $\begingroup$ @StammeringMathematician $x^2(x^2+1)$ has a real root on the interval $[-1, 1]$. Then the exact same argument as the one you use in your answer will "prove" that there must be another root (there are three remaining roots, and we can't have that none of them are real), but there isn't one. $\endgroup$ – Arthur Apr 29 '19 at 15:35
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    $\begingroup$ Whether we count multiple roots or not is technically up to the original problem author. Conventions vary. But yes, I would think that the intended interpretation is to count distinct roots. If the intended interpretation is to count with multiplicity, then you don't even need to look for roots in the first place. You just know that there can only be $0, 2$ or $4$ non-real roots, and therefore there cannot be exactly one real root. That's a bit boring, and it would seem rather pointless to give a concrete polynomial to work with. $\endgroup$ – Arthur Apr 29 '19 at 15:39

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