Multiplicities of zeros of the polynomial $p(z)=1+2z^4+\frac 7 {10}z^{10}$ 
How many zeros does $p(z)=1+2z^4+\frac{7}{10}z^{10}$ have in the unit disc ? Determine multiplicities of these zeros.

Let , $f(z)=2z^4$ and $g(z)=1+\frac{7}{10}z^{10}$.
Then, on $|z|=1$ ,  $|g(z)|\le 1+\frac{7}{10}|z|^{10}<2=|f(z)|$.
So, by Rouche's theorem , $p(z)$ has $4$ zeros in $|z|<1$. 

Trouble to find multiplicities. 

Suppose $\alpha$ be a zero  of $p(z)$ in $|z|<1$ of multiplicity $2$.
Then , $p(\alpha)=0$ and $p'(\alpha)=0$ which implies, $1+2\alpha^4+\frac{7}{10}\alpha^{10}=0$         .....$(1)$
and $8\alpha^3+7\alpha^{9}=0$.
So ,  $\alpha^6=-8/7$
If we can show that  this value of $\alpha$ does not satisfy equation $(1)$ then can we say that all roots are distinct ?
$$OR$$Any other way to show whether the roots are distinct or NOT.
 A: Answering the newer version with $p(z)=1+2z^4+\dfrac7{10}z^{10}$.
A general fact (valid in all fields, not just that of complex numbers) is that a zero of multiplicity $>1$ is a common zero of both $p(z)$ and $p'(z)$. IOW it is a zero of the greatest common divisor $d(z):=\gcd(p(z),p'(z))$.
That GCD can be very efficiently calculated with Euclid's algorithm (for the purposes of this question Wikipedia on polynomial GCD is more useful than the one explaining Euclid's algorithm for calculating the GCD of integers - algebraically they have a common root).
Here $p'(z)=8z^3+7z^9$. Any common factor of $p$ and $p'$ is also a factor of
$$
r_1=p-\frac 1{10}z\ p'=1+z^4(2-\frac8{10})+z^{10}(\frac7{10}-\frac7{10})=1+\frac65z^4.
$$
I'm sure that you can find the zeros of $r_1(z)$, and test whether they can be zeros of $p(z)$. If not, you can continue, and calculate the remainder $r_2$ of the long division of $p'$ by $r_1$. If $r_2\neq0$, then it will be of a lower degree than $r_1$, and you can use that instead.
Alternatively you can substitute $z^4=-5/6$ to the original equation $p(z)=0$. You get a quadratic that's easy to manage.
