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I have the following question in real analysis course:
let $p(x)=x^{4}+a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}$. Prove that if $x_{0}$ is a root of P, meaning $p(x_{0})=0$ and $p'(x_{0})\neq0$, then p has at least two roots.
approach: I tried to make several things, like first of looking at $p(-x_{0})$ and denote $p(-x_{0})=\frac{m}{2}$ and so $\displaystyle\frac{m}{2}=x_{0}^{4}-a_{3}x_{0}^{3}+a_{2}x_{0}^{2}-a_{1}x_{0}+a_{0}$ and then we can use the fact that $p(x_{0})=0$ and therefore $p(x_{0})-p(-x_{0})=-\frac{m}{2}$. It doesn't get me much far tho.
I assume it has to do something with the derivative because they mentioned it. any clues how to help?
Thank you all in advance

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    $\begingroup$ Draw a picture. If $p'(x_0) > 0$ then $p$ is negative for some $x < x_0$. What do you know about $\lim_{x \to -\infty} p(x)$? $\endgroup$
    – Martin R
    Commented Feb 28 at 17:29
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    $\begingroup$ If $p(x_0) = 0$ and $p'(x_0) \ne 0$ then the graph must "cross the $x$ axis" at $x_0$ and the function either goes from positive to negative or from negative to positive. No look at that function as $x\to\infty$ and as $x \to -\infty$. How many times can/must the graph cross the $x$-axis? $\endgroup$
    – fleablood
    Commented Feb 28 at 17:31
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    $\begingroup$ Hint: the wouldn't be true for $q(x) = x^3 + a_2x^2 + a_1x + a_0$ and it wouldn't be true for $q(x)=x^5 +a_4x^4 + a_3x^3 + a_2x^2 + a_1 + a_0$. Do you think it be true for $r(x) = x^2 + ax + b$? $\endgroup$
    – fleablood
    Commented Feb 28 at 17:33
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    $\begingroup$ Btw: I noticed that you got answers for all of your previous questions, but never accepted an answer so far. If you are not aware of it: Accepting an answer is important as it both rewards posters for solving your problem and informs others that your issue is resolved. See What should I do when someone answers my question? and How does accepting an answer work? for more information. $\endgroup$
    – Martin R
    Commented Feb 28 at 17:48
  • $\begingroup$ Being a quartic, $p$ has exactly 4 roots. Do you mean $p$ has at least 2 distinct roots? 2 real roots? 2 distinct real roots? $\endgroup$
    – Calvin Lin
    Commented Feb 28 at 18:29

2 Answers 2

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It is interesting that you listed as one of the tags for this question, even though you did not mention them in the body of the question. Infinitesimals are indeed relevant here, as follows. Since the derivative at $x_0$ is nonzero, there are two cases: (a) $p'(x_0)<0$, and (b) $p'(x_0)>0$. Let's consider for example case (a). Since the derivative is negative, for all infinitesimal $\epsilon>0$ one will have $p(x_0+\epsilon)<0$. On the other hand, since the polynomial is monic, for sufficiently large $N$ one will have $p(x)>0$ when $x> N$. By the intermediate value theorem, the polynomial must have another zero between $x_0+\epsilon$ and $N$.

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  • $\begingroup$ Thank you very much $\endgroup$
    – perplexed
    Commented Feb 29 at 9:52
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    $\begingroup$ @Somuser, you're welcome. Incidentally, if you need this for a homework assignment, avoid as hell mentioning infinitesimals in your solution. Some teachers have been trained in such a way that their reaction upon seeing the word "infinitesimal" is to reach for their revolver. Just say "if $\epsilon$ is sufficiently small", etc. $\endgroup$ Commented Feb 29 at 9:54
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Let p(x) be a polynomial of exact even degree n >= 2. Assume that the highest power has a coefficient of 1, otherwise we divide by the highest coefficient. Now if we calculate the limit if p(x) for x -> +infinity and x -> -infinity, both are +infinity, so we have two different points with p(x) > 0.

Your condition shows that you have t with p(t) = 0, and p(t') < 0 for either some t' < t or t' > t. Either way the polynomial has both positive and negative values in an interval that doesn't include t, and because polynomials are continuous there must be a zero at some point other than t.

You actually don't need polynomials and converging towards infinity, all you need is that p(x) is continuous everywhere, and there are arbitrary large and arbitrary small x both with p(x) > 0, or both with p(x) < 0.

For odd degrees it doesn't work; for example for $p(x) = x$, or $p(x) = x^3 + x$, or $p(x) = x^{999} + x$.

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