Question is to check :

which of the following is sufficient condition for a polynomial

$f(x)=a_0 +a_1x+a_2x^2+\dots +a_nx^n\in \mathbb{R}[x] $ to have a root in $[0,1]$.

  • $a_0 <0$ and $a_0+a_1+a_2+\dots +a_n >0$
  • $a_0+\frac{a_1}{2}+\frac{a_2}{3}+\dots +\frac{a_n}{n+1}=0$
  • $\frac{a_0}{1.2}+\frac{a_1}{2.3}+\dots+\frac{a_n}{(n+1).(n+2)} =0$

First of all i tried by considering degree $1$ polynomial and then degree $2$ polynomial and then degree $3$ polnomial hoping to see some patern but could not make it out.

And then, I saw that $a_0= f(0)$ and $f(1)=a_0+a_1+a_2+\dots +a_n$.

So, if $f(0)<0$ and $f(1)>0$ it would be sufficient for $f$ to have root in $[0,1]$

In first case we have $a_0 <0$ i.e., $f(0)<0$ and $f(1)>1>0$.

So, first condition should be implying existence of a root in $[0,1]$

for second case, let $f(x)$ be a linear polynomial i.e., $f(x)=a_0+a_1x$

Now, $a_0+\frac{a_1}{2}=0$ implies $0\leq x=\frac{-a_0}{a_1}=\frac{1}{2}< 1$ So, this might be possibly give existence in case of linear polynomials.

Now, $\frac{a_0}{1.2}+\frac{a_1}{2.3}=0$implies $0\leq x=\frac{-a_0}{a_1}=\frac{1}{3}< 1$ So, this might be possibly give existence in case of linear polynomials.

So, for linear polynomials all the three conditions imply existence of a root in $[0,1]$.

But, i guess this can not be generalized for higher degree polynomial.

I think there should be some "neat idea" than checking for roots and all.

I am sure about first case but I have no idea how to consider the other two cases.

please provide some hints to proceed further.

  • 1
    $\begingroup$ Maybe look at bounds on roots. There are some results here en.wikipedia.org/wiki/… $\endgroup$
    – Wintermute
    Nov 6, 2013 at 12:17
  • $\begingroup$ What you say $\;f(1)\;$ is in line 9 in fact is $\;f(1)-a_0\;$... $\endgroup$
    – DonAntonio
    Nov 6, 2013 at 12:28
  • $\begingroup$ Counter example to (1): $\; x^2+x-6=(x+3)(x-2)\;$ $\endgroup$
    – DonAntonio
    Nov 6, 2013 at 12:29
  • $\begingroup$ First condition would have been always true by IVT, if only $a_0 <0$ and $ a_0 + a_1 + ...+ a_n > 0$ $\endgroup$ Nov 6, 2013 at 13:04
  • $\begingroup$ @DonAntonio i am sorry i did not understand your counter example... could you please explain... $\endgroup$
    – user87543
    Nov 6, 2013 at 13:16

3 Answers 3


For the second case consider the polynomial $$ F(x)=a_0x+\frac12a_1x^2+\frac13a_2x^3+\cdots+\frac1na_{n-1}x^{n}+\frac1{n+1}a_nx^{n+1} $$ and then use Rolle's theorem.

For the third case consider some other polynomial (which?) and then use two times Rolle's theorem.

  • $\begingroup$ I did not understand this... for rolle's theorem we should have $F(a)=F(b)$ are you using same end points i.e.,$0,1$ if so, $F(0)\neq F(1)$.. I am confused... :( $\endgroup$
    – user87543
    Nov 6, 2013 at 13:34
  • $\begingroup$ @PraphullaKoushik: But $F(0)=0$ and $F(1)=a_0+\frac{a_1}{2}+\ldots+\frac{a_n}{n+1}$ right? Why $F(0)\neq F(1)$? What is $F(1)$? $\endgroup$
    – P..
    Nov 6, 2013 at 13:36
  • $\begingroup$ ok ok, if we are assuming second condition, then $F(0)=F(1)$ and so, $F'(x)=a_0 +a_1x+\dots +a_nx^n$ (I believe you mean $\frac{1}{n+1}a_nx^{n+1}$ instead of $\frac{1}{n}a_nx^{n+1}$).. So, $F'(x)$ has a root in between $0$ and $1$ and so i am done for second condition... could you please confirm If my justification clear for this part... $\endgroup$
    – user87543
    Nov 6, 2013 at 13:41
  • $\begingroup$ @PraphullaKoushik: Oops! Now is corrected and yes you are right!!! What about the third case? $\endgroup$
    – P..
    Nov 6, 2013 at 13:43
  • $\begingroup$ I just now realized how beautiful your idea is... I could now see the polynomial for 3rd case too.. I am a bit sad why i could not see this before... May be for time being i should sing something like " when i get older, I will be stronger".. then i would be able to see some beautiful things like this... I am afraid my anxiety will not be cool down unless i write down detailed note of 3rd case here (I will see it myself writing down here and feel happy about myself:P)... $\endgroup$
    – user87543
    Nov 6, 2013 at 14:29


  • $a_0 <0$ and $a_0+a_1+a_2+\dots +a_n >0$ means that $f(0)<0$ and $f(1)>0$.

  • $a_0+\frac{a_1}{2}+\frac{a_2}{3}+\dots +\frac{a_n}{n+1}=0$ means that $F(1)=0$, where $F(x)=\int_0^x f(t) \, dt$. Note that $F'=f$ and $F(0)=0$ and recall Rolle's theorem.

Can you think of what the third condition means in this context?

  • $\begingroup$ for first condition i have checked and i am sure about my proof.. i did not get your hint for second condition.. may be i need some more time.. $\endgroup$
    – user87543
    Nov 6, 2013 at 13:31
  • $\begingroup$ @PraphullaKoushik, see my edited answer. $\endgroup$
    – lhf
    Nov 6, 2013 at 13:36
  • $\begingroup$ Thank you sir, I would now try for 3rd case... Thank You... I see both the answers at a time and confused which answer should i accept... :) $\endgroup$
    – user87543
    Nov 6, 2013 at 13:44

for third case we consider polynomial

$F(x)=\frac{a_0}{1.2}x^2+\frac{a_1}{2.3}x^3+\dots + \frac{a_n}{(n+1)(n+2)}a_nx^{n+2}$

we now assume third condition i.e., $\frac{a_0}{1.2}+\frac{a_1}{2.3}+\dots+\frac{a_n}{(n+1).(n+2)} =0$

In that case, for polynomial $F(x)$ we would then have $F(0)=0$ and $F(1)=0$ (with given condition)

So, by rolle's theorem we have a root for $F'(x)$ in $[0,1]$

i.e., we have a root for $F'(x)=\frac{a_0}{1}x+\frac{a_1}{2}x^2+\dots+ \frac{a_n}{n+1}x^{n+1}$ in $[0,1]$ say at $c\in [0,1]$

Now, for $F'(x)$ we have two zeros.. i.e., $F'(0)=0$ and $F'(c)=0$

Now, i will use rolle's theorem again i.e, i have root for $F''$ in $[0,c]$

where $F''(x)=a_0+a_1x+a_2x^2+\dots+a_nx^n$

to conclude, i now set $F''(x)=f(x)$ and with given condition,

i have a root in $[0,c] $ for some $c\in [0,1]$ particularly, it has a zero in $[0,1]$

i.e., $f(x)=a_0+a_1x+a_2x^2+\dots+a_nx^n$ has a root in $[0,1]$

To conlcude, with above answer and my previous observation of second case,

$f(x)=a_0 +a_1x+a_2x^2+\dots +a_nx^n\in \mathbb{R}[x] $ have a root in $[0,1]$ in all three following cases:

  • $a_0 <0$ and $a_0+a_1+a_2+\dots +a_n >0$
  • $a_0+\frac{a_1}{2}+\frac{a_2}{3}+\dots +\frac{a_n}{n+1}=0$
  • $\frac{a_0}{1.2}+\frac{a_1}{2.3}+\dots+\frac{a_n}{(n+1).(n+2)} =0$

P.S : This is completely for the sake of my reference and all the credit goes to above two users who have helped me to go through this idea.


You must log in to answer this question.