<s>[INST] How to find the roots of the polynomial We consider the irreducible polynomial $g=y^4+y+1 \in \mathbb{F}_2[y]$ and let $b$ be a root of $g$. 
I want to find all the roots of $g$ and also three generators of $\mathbb{F}_{16}^{\ast}$ as for the basis $\{1, b, b^2, b^3 \}$.
I have tried the following:
I applied the euclidean division of $y^4+y+1$ with $y-b$ and I got $y^4+y+1=(y-b) (y^3+by^2+b^2 y+(b^3+1))$. 
Then I applied the euclidean division of $y^3+by^2+b^2 y+(b^3+1)$ with $y-(b+1)$ and I got $y^3+by^2+b^2 y+(b^3+1)=(y-(b+1))(y^2+y+(b^2+b+1))$. 
Then suppose that the third root is $r_1$. Then applying the euclidean division of $y^2+y+(b^2+b+1)$ with $y-r_1$ we get $y^2+y+(b^2+b+1)=(y-r_1)(y+(1+r_1))$ when $r_1^2+r_1+b^2+b+1=0$. 
How can we solve the last equation to get $r_1$? Can we use the discriminant? 
 [/INST] Observe that if $ b $ is a root of $ g $, then so are $ b+1, b^2, b^2 + 1 $. All of these are obviously pairwise distinct since $ g $ is an irreducible polynomial.
EDIT : $ b^2 $ is a root because we have $$ g(b^2) = (b^2)^4 + b^2 + 1 = b^8 + b^2 + 1 = (b+1)^2 + b^2 + 1 = 0 $$ From the fact that $ b $ is a root $ \implies $ $ b + 1 $ is a root, we get that $ b^2 + 1 $ is also a root.  
To answer your second question, you want to find a generator for $ \mathbb{F}_{16}^{\times} \cong \mathbb{F}_2(b)^{\times} $. We know that $ b^{15} = 1 $. Further, $ b^3 \neq 1 $, $ b^5 = b^2 + b \neq 1 $, so the order of $ b $ is indeed $ 15 $ and hence it is a generator for the multiplicative group. How then, can you find all other generators?
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