# conditions imposed on a polynomial for its roots

Consider $P(x)=a_0+a_1x+a_2x^2+\cdot\cdot\cdot\cdot\cdot\cdot +a_nx^n,a_i's \in \mathbb R$.The question is whether the following conditions can guarantee existence of a root of $P(x)$ in $[0,1]$?

a.$a_0<0$ and $a_0+a_1+ \cdot\cdot\cdot\cdot\cdot+a_n>0$

b.$a_0+a_1/2+a_2/3+\cdot\cdot\cdot\cdot\cdot\cdot +a_n/(n+1)=0$

c.$a_0/1.2+a_1/2.3+\cdot\cdot\cdot\cdot\cdot\cdot +a_n/(n+1)(n+2)=0$

(a) is true since $P(0)<0$ and $P(1)>0$.but how to do other two?

Hint for (b): calculate $\int_0^1 P(x) \, dx$ and use the fact that if $P$ doesn't vanish on $[0,1]$ then it's always positive or always negative.

Hint for (c): generalize (b).

• That's a question for you. Try to see first why it's true intuitively, and then see if you can remember a theorem that will help you prove it formally. – Yuval Filmus Nov 1 '14 at 7:19
• Well sorry for my belated reply @Yuval.Please see my try.If $P(x)$ does not vanish and $P(x)>0$ for some $x$ and $P(x)<0$ then $P(x)$ being continuous by IVP $P(x)=0$ for some $x$ hence will vanish.Then either $P(x)>0 \forall x$ or $P(x)<0 \forall x$.If that is true $\int _{0}^{1}P(x)$is either >0 or <0 again contradiction.thus $P(x)$ will vanish in $[0,1]$ – Learnmore Nov 1 '14 at 8:19
• Right, that's the idea. – Yuval Filmus Nov 1 '14 at 18:04