<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] If the Descartes rule does not help there is always the following (or similar) way: $$x^7+5x^5+x^3−3x^2+3x−7=(x-1)(x^6+x^5+6x^4+6x^3+7x^2+4x+7).$$
But $$x^6+x^5+6x^4+6x^3+7x^2+4x+7=$$
$$=\left(x^3+\frac{1}{2}x^2-1\right)^2+\frac{1}{4}(23x^4+40x^3+36x^2+16x+12)=$$
$$=\left(x^3+\frac{1}{2}x^2-1\right)^2+\frac{1}{4}(23x^4+40x^3+18x^2+18x^2+16x+12)>0,$$
which says that our equation has an unique real root.
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