# Pick out the case(s) which ensure that the polynomial $p(\cdot)$ has a root in the interval $[0, 1]$ [duplicate]

This question already has an answer here:

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)$\frac{a_0}{1.2}+\frac{a_1}{2.3}+ \dots+\frac{a_n}{(n + 1)(n + 2)}= 0$.

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

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

(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

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