# 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?

• $x=1$ is already a root – Claude Leibovici Sep 18 at 7:30
• Factor $(x-1) \left(x^6+x^5+6 x^4+6 x^3+7 x^2+4 x+7\right)=0$ – Raffaele Sep 18 at 7:33

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