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?

Descare's rukles say that there will be at most 3 positive roots and no negative root. Since $f(1)=0$, so $(x-1)$ is factor of $f(x)=x*7+5x^5+x^3-3x^2+3x-7$
Note that $f(x)=(x-1)(x^6+x^5+6x^3+7x^2+4x+7)=(x-1)g(x)$, It can be seen that
the number of sign changes in $g(x)$ are none, so $g(x)$ will have no positive roots. Finally, $f(x)=0$ will have exactly one real root, namely $x=1$.
