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This question already has an answer here:

Please help me to solve the problem below.

Consider the polynomial $p(x) = a_0 + a_1x + a_2x^2+\dots+ a_nx^n$, with real coefficients.
Pick out the case(s) which ensure that the polynomial $p(\cdot)$ has a root in the interval $[0, 1]$.

(a) $a_0 < 0$ and $a_0 + a_1 +\dots+ a_n > 0$.
(b) $a_0 +a_1/2+ \dots +a_n/(n + 1)= 0$.
(c)$\frac{a_0}{1.2}+\frac{a_1}{2.3}+ \dots+\frac{a_n}{(n + 1)(n + 2)}= 0$.


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marked as duplicate by Arnaud D., Xander Henderson, Theoretical Economist, max_zorn, José Carlos Santos Sep 5 '18 at 9:13

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Do you know results which might help, like the intermediate value theorem and Rolle's theorem? $\endgroup$ – Antonio Vargas Jan 4 '13 at 14:26
  • $\begingroup$ The linked question came later, but I think it is better, and has better answers. $\endgroup$ – Arnaud D. Sep 4 '18 at 12:01
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(b)

$q(x)$ = $a_0 \cdot x +a_1\cdot \frac{x^2}{2} +....+a_n\cdot\frac{x^n+1}{n + 1} $

$q(0)=0=q(1)$ then by rolles theorem $ \implies p(x) = 0$ ( $p(x)$ is the derivative of $q(x)$)

similar you can approach (c) also

(a)

$p(0) =a_0 < 0$ and $p(1)>0$ then use intermediate value theorem

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  • $\begingroup$ I think you mean intermediate value theorem for part (a). $\endgroup$ – Antonio Vargas Jan 4 '13 at 14:37
  • $\begingroup$ @AntonioVargas sorry for the mistake i have corrected it $\endgroup$ – jim Jan 4 '13 at 14:45

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