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For a lower degree polynomial function, we can simply expand the product of factors using distributive properties and compare that to the original polynomial. But how to prove that an arbitary degree single variable polynomail indeed equals to product of x minus all its zeros (complex numbers), in another word, validatity of factoring?

techie11
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    You use the Factor Theorem to prove that $x-a$ divides $p(x)$ if and only if $p(a)=0$. Then you use induction and the Fundamental Theorem of Algebra to show you can write it that way. Then you use unique factorization to prove that it can be written in only that way. – Arturo Magidin Jan 21 '20 at 20:43
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    @ArturoMagidin Perfectly explained. Why not post an answer ? – Peter Jan 21 '20 at 20:54
  • Two slight remarks : $(1)$ The degree of the polynomial must be $1$ or greater (no constant polynomial) $(2)$ The roots can have multiplicity greater than $1$. Considering the multiplicity , a polynomial with degree $n$ has exactly $n$ complex roots and therefore splits into linear factors over $\mathbb C[x]$. Such a field is called algebraically closed. – Peter Jan 21 '20 at 20:56
  • ArturoMagidi, Peter: Thank you very much – techie11 Jan 21 '20 at 21:36

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First we prove (Equality 1), for all $x$ $$f(x)=(x-a)\cdot g(x)+b$$ With $a,b$ any number, $f(x)$ a polinomial function of degree $n$ and $g(x)$ polynomial function of degree $k\leq n$. The proof will be done by complete induction on $n$.

(1) When $n=1$, $$px+q=px-pa+q+pa=(x-a)p+(q+pa)$$ for $g(x)=p\land b=q+pa$ $$=(x-a)\cdot g(x)+b$$ It's true for $n=1$

(2) The equality is assumed to be true for $k\leq n$.

For $n=k+1$ $$f(x)=a_{k+1}x^{k+1}+\ldots +a_{1}x+a_0$$ So that $$h(x)=f(x)-a_{k+1}(x-a)$$ $h(x)$ has a degree $\leq k\leq n$ so, by our assumption, $$h(x)=f(x)-a_{k+1}(x-a)=(x-a)g(x)+b$$ $$f(x)=(x-a)(g(x)+a_{k+1})+b$$ (remember that $f(x)$ has a degree of $k+1$ so we proved that if its true for $k$ it´s true for $k+1$.) $\square$

Equality 1 gives us $f(x)=(x-a)\cdot g(x)+b$, but since $g(x)$ is also a polinomial, $$f(x)=(x-a)\big((x-c)j(x)+r\big)+b=(x-a)(x-c)j(x)+(x-a)r+b$$ The same goes for $j(x)$ and so on. $$f(x)=(x-a)(x-c)(x-d)\ldots +\text{a bunch of stuff (rigorous AF)}$$ When $x=$ one of the zeroes: $$0=f(x)=0+\text{a bunch of stuff}\Rightarrow \text{a bunch of stuff}=0$$ So, when $x=$ one of the zeroes $$f(x)=(x-a)(x-c)(x-d)\ldots$$ $\square$

Sebastián P. Pincheira
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