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?
 A: 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.
A: 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.
A: 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 $x=1$.
A: 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: $D)1$
