<s>[INST] Proof of Different Polynomial Decompositions into Linear Factors From G. Polya "Mathematics and Plausible Reasoning" p. 18.
How do you prove that provided the roots of a polynomial are different from zero, 
$$a_0 + a_1x+a_2x^2 + ... + a_nx^n$$
$$\,= a_0\left(1-\frac{x}{\alpha_1}\right)\left(1-\frac{x}{\alpha_2}\right)...\left(1-\frac{x}{\alpha_n}\right)$$
with $\alpha_1, \alpha_2,...\alpha_n$ corresponding to the polynomial roots.
 [/INST] \begin{align}
&a_0+a_1x+\ldots+a_nx^n=\\
&\qquad=a_n(x-\alpha_1)\cdot\ldots\cdot(x-\alpha_n)=\\
&\qquad=a_n\alpha_1\frac{x-\alpha_1}{\alpha_1}\cdot\ldots\cdot\alpha_n\frac{x-\alpha_n}{\alpha_n}=\\
&\qquad=a_n\alpha_1\cdot\ldots\cdot\alpha_n\left(\frac{x}{\alpha_1}-1\right)\cdot\ldots\cdot\left(\frac{x}{\alpha_n}-1\right)=\\
&\qquad=(-1)^n\,a_n\alpha_1\cdot\ldots\cdot\alpha_n\left(1-\frac{x}{\alpha_1}\right)\cdot\ldots\cdot\left(1-\frac{x}{\alpha_n}\right)
\end{align}
Setting $x=0$ results in:
$\qquad a_0+a_1x+\ldots+a_nx^n\,=\,(-1)^n\,a_n\alpha_1\cdot\ldots\cdot\alpha_n\left(1-\frac{x}{\alpha_1}\right)\cdot\ldots\cdot\left(1-\frac{x}{\alpha_n}\right)$
simplifying to 
$$a_0=(-1)^n\,a_n\alpha_1\cdot\ldots\cdot\alpha_n$$ Therefore:
$\qquad a_0+a_1x+\ldots+a_nx^n \,=\,a_0\left(1-\frac{x}{\alpha_1}\right)\cdot\ldots\cdot\left(1-\frac{x}{\alpha_n}\right)$
</s>