How does the fundamental theorem of algebra follow from Weierstrass’s theorem. Can anyone please explain to me how the fundamental theorem of
algebra follows from or is related to the Bolzano Weierstrass’s theorem?
 A: As mentioned in the comments, there is a great connection between the Bolzano-Weierstrass theorem (BWT) and the fundamental theorem of algebra (FTA). I will go over this paper and try to break it down. First, to state the theorems:
BWT: Every bounded sequence in $\mathbb{R}^n$ has a convergent subsequence. Note that for the purposes below, it can be stated that any continuous function $f:\mathbb{D}\to\mathbb{R}$ (where $\mathbb{D}$ is a closed and bounded disk) has a minimum on $\mathbb{D}$.
FTA: Every non-constant single-variable polynomial with complex coefficients has a minimum of one complex root.
This paper gives a proof of the FTA via BWT. I will go through the technical details if you like, but since you asked, I will give an overview of what was written. 
The reason the BWT is used is because it avoids the use transcendental functions (analytic function that does not satisfy a polynomial equation) since their theory is far deeper than the FTA. 
For the proof, the first step is to establish an inequality that bounds $0$ above and below by the imaginary and real parts of a given $\zeta$ to some power greater than or equal to $2$. THe power is a natural number excluding the origin. The proof is via the Binomial formula.
The idea of the proof is to introduce a polynomial with coefficients in the complex plane. Call this $P$. Multiply this $P$ with its conjugate $\overline{P}$. Take the limit as $|z|\to\infty$ to establish that $P\overline{P}$ has a global minimum at some point $z_0\in\mathbb{C}$. 
By assuming $z_0=0$ we can write $P\overline{P}$ minus the same quantity evaluated at zero (which is clearly positive). Using the substitution $r\zeta$ (where $\zeta$ is a complex number and a positive scalar $r$) and analysing the cases for $k$ odd and even separately finished the proof.  
Is this okay? 
