Sum And Product Of The Roots

Theorem: If $f(x)=a_{n}x^n + a_{n-1}x^{n-1} + \cdots + a_{0}x^0$ is a polynomial,$\\[12pt]$ then the normalized coefficients with alternate '$+$'and '$-$' signs, i.e., $-\dfrac{a_{n-1}}{a_{n}}, \dfrac{a_{n-2}}{a_{n}}, \cdots , (-1)^n\dfrac{a_{0}}{a_{n}}$ gives the different sums and products of the roots of the equation $f(x)=0$.

Prerequisites:
Factor Representation of a Polynomial (proof)

Proof:
Let the given polynomial be:

$\qquad\quad f(x)=a_{n}x^n + a_{n-1}x^{n-1} + \cdots + a_{0}x^0  \qquad\qquad\qquad\qquad\qquad\qquad\qquad\cdots (1)$

This can be written in the form of factor representation as:

$\qquad\quad f(x) = (x-\alpha_1)(x-\alpha_2) \cdots (x-\alpha_n) g(x)\qquad\qquad\qquad\qquad\qquad\qquad\;\;\cdots (2)$

Here $\;\alpha_1, \alpha_2, \cdots ,\alpha_n$ are the roots of the equation $f(x)=0$.
Also, $\;deg\big(g(x)\big) = n-n = 0$. This implies that $g(x)$ is a constant $= g$ (say).

Expanding equation $(2)$:

$\qquad\quad f(x) = gx^n - g\left(\alpha_1+\alpha_2+\cdots +\alpha_n\right)x^{n-1} + g\left(\alpha_1\alpha_2+\alpha_2\alpha_3+\cdots +\alpha_{n-1}\alpha_n\right)x^{n-2}\\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad + \cdots + (-1)^n g\left(\alpha_1\alpha_2\cdots\alpha_n\right)$

Comparing coefficients with equation $(1)$,

$\qquad\quad g = \alpha_n\\ \qquad\quad\displaystyle\sum\limits_{i=1}^n \alpha_i = \alpha_1 +\alpha_2+\cdots +\alpha_n = - \dfrac{a_{n-1}}{a_{n}}\\ \\ \qquad\quad\displaystyle\sum\limits_{i\neq j} \alpha_i\alpha_j =\alpha_1\alpha_2+\alpha_2\alpha_3+\cdots +\alpha_{n-1}\alpha_n = \dfrac{a_{n-2}}{a_{n}}\\ \\ \qquad\quad\;\;\vdots\\[6pt] \\ \qquad\quad\displaystyle\prod\limits_{i=1}^n \alpha_i = \alpha_1\alpha_2\cdots\alpha_n = (-1)^n\dfrac{a_{0}}{a_{n}}$

Hence the result.


Recommended:
Condition for roots of quadratic equation
Quadratic Formula
Factor Theorem