<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] The first test of factors using the rational root theorem yields
$$(x - 1) (x^6 + x^5 + 6 x^4 + 6 x^3 + 7 x^2 + 4 x + 7) = 0$$
From this we can see that, for one root, $X=1$.
From the co-factor, we can see that the function never crosses the $x$-axis so there are no more real roots.
Answer: $D)1$
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