Given the polynomial $P(x) = x^7 + x^6 - 7x^5 - 7x^4 + 8x^3 + 8x^2 + 16x + 16$, we know that its Sturm's sequence is given by $$\{P(x), P_1(x), P_2(x), P_3(x), P_4(x), P_5(x)\}$$ with $P_5(x) = -x^2 + 4$. Determine all roots of P.
My attempt
First of all, it is important to be said that this question appeared in a math exam where no software computation nor calculators where allowed. Having said that, I approached the problem the following way:
First of all $P_1 = P'(x) = 7 x^6 + 6 x^5 - 35 x^4 - 28 x^3 + 24 x^2 + 16 x + 16$.
We can then take $P_2 = -104 x^5 -112 x^4 + 252 x^3 + 256 x^2 + 656 x + 768 = -26 x^5 - 28 x^4 + 63 x^3 + 64 x^2 + 164 x + 192$
We can then take $P_3 = -113 x^4 - 100 x^3 + 444 x^2 + 400 x + 32$.
We can then take $P_4 =-35 x^3 - 48 x^2 + 140 x + 192$ and we already knew that $P_5 = -x^2 + 4$. We can make the signs table:
From here, we know that there are $4-0=4$ positive roots, but we get that there are $0-1=-1$ negative roots, which does not make sense at all.
I may have made a mistake somewhere, so any help is truly appreciated. I also think that I may be going in the wrong direction since the divisions are sort of hard for not even having a calculator in the exam.
Just by checking one can see that $-1$ is a root of $P$, i.e., $P(x)=(x+1)(x^6-7x^4+8x^2+16)$. The polynomial $x^6-7x^4+8x^2+16$ has symmetric roots in the sense that if $a$ is a root, then so is $-a$. One now readily notices that $2$ is a root and hence so is $-2$. Thus $P(x)=(x+1)(x-2)(x+2)(x^4-3x^2-4)$. From your computation is follows that $P'(2)=P'(-2)=0$, i.e., $\pm 2$ is a double root of $P(x)$. Thus we finally arrive at $P(x)=(x+1)(x-2)^2(x+2)^2(x^2+1)$ and we can say that the roots of $P$ are $-1,\pm 2,\pm i$.