# What is the number of real roots of the polynomial

The number of distinct real roots of the equation

$x^9+x^7+x^5+x^3+x+1=0.$

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What have you tried? Do you know what the graph of this function looks like? – Mark Bennet Aug 23 '12 at 6:16
no Sir....But I want to solve it without using graph. – ram Aug 23 '12 at 6:20
$$1+x+x^3+x^5+x^7+x^9=\left(\frac{x^{10}-1}{x^2-1}\right)x+1$$ might be helpful... – J. M. Aug 23 '12 at 6:25
I ask about whether you know anything about the graph, because this is the way I grew my knowledge of how functions work. There is some sophisticated machinery for finding the number of roots in a general case. Here you are adding simple powers of $x$ - and, for example, $x^9$ is an increasing function. What happens when you add increasing functions together? – Mark Bennet Aug 23 '12 at 6:34
plotting this will indicate -0.620374 is the root if you need it for practical purposes. – user38537 Aug 23 '12 at 13:28

Let $f(x)=x^9+x^7+x^5+x^3+x+1$, then $f^\prime (x)=9x^8+7x^6+5x^4+3x^2+1\geq1>0.$ Hence $f$ is strictly increasing, so has atmost one real root. Also $f(0)=1>0$ and $f(-1)=-4<0$ and so has a root between $-1$ and $0$.