<s>[INST] Find how many roots of $z^5+2z^2+7z+1=0$ lie inside $|z|=2$? How many are real? Find how many roots of $z^5+2z^2+7z+1=0$ lie inside $|z|=2$? How many are real?
Let $f(z)=z^5+2z^2+7z+1$, and $g(z)=z^5$. For $|z|=2,$ we note that
$$
|f(z)-g(z)|\leq2|z|^2+7|z|+1=23<32=2^5=|g(z)|.
$$
Since $g$ has $5$ zeros in $|z|\leq 2$, Rouche's theorem implies that $f$ also has $5$ zeros lie inside $|z|=2$. 
However, I have no idea to find how many of them are real. Can anyone tell me how to do it? Thanks.  
 [/INST] I'd never seen Sturm's theorem before, but it was interesting enough that I'll add the computation:
\begin{align}
p_0(x) &= x^5+2x^2+7x+1\\
p_0'(x) = p_1(x) &= 5x^4+4x+7\\
\text{remainder of $-p_0/p_1$:}\quad p_2(x) &= -\tfrac{6}{5}x^2 - \tfrac{28}{5}x - 1\\
\text{remainder of $-p_1/p_2$:}\quad p_3(x) &= \tfrac{12562}{27}x + \tfrac{8669}{108}\\
\text{remainder of $-p_2/p_3$:}\quad p_4(x) &= -\tfrac{87841935}{1262430752}\\
\end{align}
Computing at the end points, 
$$p_0(-2)=-37,\;p_1(-2)=79,\;p_2(-2)=\tfrac{27}{5},\;p_3(-2)=-\tfrac{3401}{4},\;p_4(-2)=-\tfrac{87841935}{1262430752}$$
$$p_0(2)=55,\;p_1(2)=95,\;p_2(2)=-17,\;p_3(2)=\tfrac{109165}{108},\;p_4(2)=-\tfrac{87841935}{1262430752}$$
Hence at $-2$ we have the sign sequence $-++--$, which has $2$ sign changes and at $2$ we have $++-+-$, which has $3$ sign changes and so there are $3-2=1$ real roots in $[-2,2]$.
Alternatively, we may note that the second derivative $4(5x^3+1)$ is zero at $x = -\tfrac{1}{\sqrt[3]{5}}$, which is a global minimum of the derivative, which is positive there, hence $x^5+2x^2+7x+1$ is monotone increasing everywhere. As $p_0(-2) < 0 < p_0(2)$, by continuity of polynomials and intermediate value theorem there is precisely one root in $[-2,2]$.
</s>