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Unsure on the procedure on this one and then how to explain it. I don't think this function has any rational roots, right?

  • 3
    It depends on what you're allowed to use. If you evaluate $f(-2)$ and $f(2)$, what do you get? If you're allowed something like the [intermediate value theorem](http://en.wikipedia.org/wiki/Intermediate_value_theorem) and you know that polynomials are continuous, then these two values get you what you want. At the level of precalculus I don't know offhand.2011-12-15

5 Answers 5

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Hint:

$f(-2) = (-2)^5 -2(-2) + 10 = -32 + 4 + 10 = -18 < 0$ while $f(2) = 2^5 - 2(2) + 10 = 38 > 0.$

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You can use Sturm's Theorem and Descartes's Rule of Signs.

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As I understand, you actually have three questions:

  • Does $f(x)=x^5-2x+10$ has zeros on the interval $[-2,2]$? (Notice that you can either say the root of $f(x)=0$ or zeros of $f(x)$. I don't think people would say "the root of $f(x)$".)
  • How to prove the existence of non-existence above?
  • If the root of $f(x)=0$ exists, is it rational?

Here are my answers:

  • First, try it on Mathematica. You can see the answer from the picture.

enter image description here

  • 3
    I think saying "roots of polynomials" is very common, actually.2011-12-15
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Calculate $f(2)$ and $f(-2)$. In your case they are 38 and -18 respectively. Since $f(x)$ changes its sign as one decreases $x$ from 2 to -2, $f(x)$ must have crossed the $f(x)=0$ line at some $x$. This proves that $f(x)$ has a root somewhere in the interval $[-2,2]$.

PS : Given the fact that $f(x)$ is continuous. See a comment below.

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    Yes, I totally agree. Thanks! Sorry for leaving a gap!2011-12-15
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If you want to know if a polinomial has rational roots, you use the Rational Roots Theorem. For a polynomial $a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0$ All rational roots must be of the form $\pm \frac{p}{q}$ where $p$ is a factor of $a_0$ and $q$ is a factor of $a_n$. No other rational roots may exist. In this case, you have $p=\{1,2,5,10\}$ and $q=1$, so your possible rational roots are $\{\pm1,\pm2,\pm5,\pm10\}$. If you evaluate the function at these points (don't do it by hand), you'll see none of them equal zero, so your function has no rational roots.