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Problem
The polynomial $x^5 -10x + 20$ has
a. both positive and negative real roots
b. only positive real roots
c. only negative real roots
d. at least two complex roots

My Approach

  • Tried to solve this using Descartes Rule of Sign Change and its correspondence to real roots.
  • In this case, for $f(x)$ the change of signs is $2$ so positive real roots is $\leq2$ and for $f(-x)$ the change of signs is $1$ so negative real roots $\leq1$ (or one negative real root).
  • Hence, the complex roots$\geq2$. So option d is correct.

Though the answer key listed option c as Correct Answer. Could someone help me out where I am going wrong or probably some other approach to solve this question?

Rohit Singh
  • 1,143
Ramesh Sharma
  • 29
  • Think about the local extrema – Claude Leibovici May 17 '22 at 12:16
  • Have a look at the example $x^5-10x+35$, and how it is done. The case here is almost the same. We have a negative real root and four nonreal roots. – Dietrich Burde May 17 '22 at 12:17
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    Your answer is correct. – moqui May 17 '22 at 12:36
  • There is just 1 real root, and it is negative. Let $f(x)=x^5-10x+20,.$ Then $f'(x)=5(x^4-2).$ So for real $x$ we have $|x|>2^{1/4}\implies f'(x)>0$ and $|x|<2^{1/4}\implies f'(x)<0. $ So $f(x)$ has local extrema only at $x=\pm 2^{1/4}.$ And $f(\pm 2^{1/4})>0.$ – DanielWainfleet May 17 '22 at 13:25
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    Both (c) and (d) are correct (assuming "only negative real roots" is interpreted as meaning that the only real roots are negative, not that the only roots are negative reals). If the answer guide allows only (c), it is wrong. – Robert Israel May 17 '22 at 13:57
  • @RobertIsrael I guess if we go into the language, then option c has word "roots" implying multiple root while here we can have only one negative real root.That assumption sounds unconventional to me.Though thanks. – Ramesh Sharma May 17 '22 at 13:59

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