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This question already has an answer here:

Why do polynomial of degree x have x zeroes? I searched on the net but could not find anything. Is there a derivation or proof?

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marked as duplicate by Matthew Leingang, Parcly Taxel, miracle173, kingW3, Simply Beautiful Art Sep 24 '17 at 12:08

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Fundamental theorem of algebra (1),

$ \forall P\in\mathbb{C}[X] ,\deg(P)>0 $ have at least one root in $\mathbb{C}$

with this second proposition (2) :

$r \in \mathbb{C}$ is a root of $P$ if and only if $(X-r)|P$

Let $P\in\mathbb{C}[X]$ and $\deg(P)=n$

With the first theorem (1) there exists one root in $\mathbb{C} : z_1$

So $(X-z_1)|P\quad (2) \iff P=(X-z_1)Q_1$

We apply the same method for $Q_1$ $\qquad(\deg(Q_1)=n-1)$

There exists $z_2$....

and by iteration you will find $n$ roots

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