Any irreducible polynomial in $\mathbb F_2[x] $ of degree $5$ has distinct roots in any algebraic closure of $\mathbb F_2$. Here $\mathbb F_2$ is the $2$ elements field.
My logic is we know that $\mathbb F_2 $ is a perfect field, hence every algebraic extension is separable, so every algebraic closure is separable,hence the polynomial has distinct roots.
Is my answer right ? Any insight. Thank you.
Your argument is correct.
Without knowing that finite fields are perfect, the case at hand can be shown directly:
Let $f\in\Bbb F_2[x]$ be irreducible of degree $5$ and assume $f$ has some multiple root $\alpha$ in $\overline{\Bbb F_2}$. Then $$f(x) = x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+1.$$ with $a_i\in\Bbb F_2$. Note that $f(x+1)=x^5+(a_4+1)x^4+\ldots$ is also irreducible of degree $5$ and has $\alpha+1$ as multiple root in $\overline{\Bbb F_2}$. Therefore, we may assume wlog. that $a_4=0$.
Note that the multiple root $\alpha$ is also a root of the formal derivative $f'(x)= x^4+a_3x^3+a_1x$ and therefore also of $g(x)=f(x)-xf'(x)=a_2x^2+1$.
As $g(\alpha)=0$, we cannot have $a_2=0$. But then $g(x)=(x+1)^2$ and we conclude $\alpha=1$, contradicting irreducibility of $f$.