Real roots of $x^7+5x^5+x^3−3x^2+3x−7=0$? The number of real solutions of the equation,
$$x^7+5x^5+x^3−3x^2+3x−7=0$$
is
$$(A) 5 \quad (B) 7 \quad (C) 3 \quad (D) 1.$$
Using Descartes rule we may have maximum no. of positive real roots is $3$ and negative real root is $0.$ So there can be either $3$ real roots or $1$ real root but how to conclude what will be the no. of real roots exactly. Can you please help me?

It is easy, by the rational root theorem, to find that $1$ is the only rational root. Then, dividing the polynomial by $x-1$ you get
$$
x^6+x^5+6 x^4+6 x^3+7 x^2+4 x+7
$$
that, by Decartes Rule, has zero positive roots.
