Find the number of real roots of given equation. 
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*$x^9+x^7+x^5+x^3+x+1=0$

*$x^3+x-1=0$


I know that the complex roots exist in pairs and also that the number of distinct roots has to do something with change in sign of the equation, but how?
How should i proceed in such type of questions?

for the equation $x^9+x^7+x^5+x^3+x+1=0$ we take the function f(x)=$x^9+x^7+x^5+x^3+x+1$...now see f ' (x)=$9x^8+7x^6+5x^4+3x^2+1$ >0 for all x$\in$R. So f(x) is stricktly increasing for all x$\in$R...so it wil obviously cut the X axis at exactly one point.so the equation will have only one real root as f(x)  is continuous for all x$\in$R ( as f(0)=1>0 and f(-1)=-4<0 )


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*similarly for $x^3+x-1=0$ f(x)=$x^3+x+1$ and f ' (x)=$3x^2+1$ >0 for all x$\in$R so as  f(0)=-1 and f(1)=1 so f(x) will have a root between x=0 and x=1 and exactly one real root.

