<s>[INST] 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?
 [/INST] 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.
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