Relation betwen coefficients and roots of a polynomial 
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Create polynomial coefficients from its roots 

I am reading the first chapter titled Numerical Solutions Of Equations And Interpolation by K.A. Stroud (Advanced Engineering Math) page 4. This is how it goes
Let $\alpha$,$\beta$,$\gamma$ be the roots of $x^3 + px^2 + qx + r = 0$. Then,writing the expression $x^3 + px^2 + qx + r$ in terms of $\alpha$, $\beta$ and $\gamma$ gives $(x - \alpha)(x - \beta)(x - \gamma)$.
$\therefore$ $x^3 + px^2 + qx + r = (x - \alpha)(x - \beta)(x - \gamma)$.
$= (x^2 - [\alpha + \beta]x + \alpha\beta)(x - \gamma)$
$= x^3 - (\alpha + \beta)x^2 + \alpha\beta x - \gamma x^2 + (\alpha + \beta)\gamma x - \alpha\beta\gamma$
$= x^3 - (\alpha + \beta + \gamma) x^2 + (\alpha\beta + \beta\gamma + \gamma\alpha) x - \alpha\beta\gamma$
$\therefore$ equating coefficients
(a) $\alpha + \beta + \gamma = -p$.
(b) $\alpha\beta + \beta\gamma + \gamma\alpha = q$.
(c) $\alpha\beta\gamma = -r$.
This, of course, applies to a cubic equation. Let us extend this to a more general equation
In general, if $\alpha_{1}$, $\alpha_{2}$, $\alpha_{3},\ldots,\alpha_{n}$ are the roots of the equation $p_{0}x^n + p_{1} x^{n-1} + p_{2}x^{n-2} + \cdots + p_{n-1}x + p_{n} = 0$ where ($p_{0}\neq0$) then 
sum of the roots $= \frac{-p_{1}}{p_{0}}$
sum of products of the roots, two at a time $= \frac{p_{2}}{p_{0}}$
sum of products if the roots, three at a time $= \frac{-p_{3}}{p_{0}}$
sum of products of the roots, $n$ at a time = $(-1)^n,\frac{p_{n}}{p}_{0}$
I was able to understand the cubic equation's part but I am completely lost with the general part (i.e an $n$th degree polynomial). I am looking for a simpler explanation of what that means. However I understand that it can be used as formulas for finding the roots but i need to know how did he obtain the above formulas.
Thanks in advance.
 A: The Factor Theorem says that if a polynomial $p(x)$ has root $r$, then $x-r$ divides $p(x)$. 
Since polynomial with complex coefficients always have exactly the same number of roots as its degree (counting multiplicity), and they have unique factorization, that means that if $p(x)$ is the polynomial
$$x^n + p_1x^{n-1} + \cdots + p_{n-1}x + p_0$$
(note that I am doing the monic case first; I'll deal with the general case later) and $r_1,\ldots,r_n$ are the roots of $p(x)$, then we can factor $p(x)$ as
$$p(x) = x^n + p_1x^{n-1}+\cdots + p_{n-1}x + p_0 = (x-r_1)(x-r_2)\cdots(x-r_n).$$
When you multiply out  $(x-r_1)\cdots(x-r_n)$, and then put together the powers of $x$, you get a polynomial expression in which the coefficients are written in terms of the roots. For example, to get the coefficient of $x$, note that you will get an $x$ when you expand and multiply the $x$ in the first binomial by each of the constants in the rest, giving you $(-1)^{n-1}(r_2\cdots r_n)x$. Another when you multiply the $x$ in the second binomial by each of the constants in the rest; another when you multiply the $x$ in the third binomial by the constants in the rest; etc.
If you work this out, you'll find that when you expand $(x-r_1)\cdots(x-r_n)$ and then group together the powers of $x$, you will have:


*

*The coefficient of $x^n$ is $1$.

*The coefficient of $x^{n-1}$ is $-(r_1+\cdots + r_n)$.

*The coefficient of $x^{n-2}$ is $$r_1r_2 + r_1r_3 + \cdots +r_1r_n + r_2r_3+\cdots + r_{n-1}r_n.$$

*The coefficient of $x^{n-3}$ is $$-\bigl(r_1r_2r_3 + r_1r_2r_4+\cdots + r_1r_2r_n + r_1r_3r_4 + \cdots + r_{n-2}r_{n-1}r_n\bigr).$$

*$\vdots$

*The coefficient of $x$ is $$(-1)^{n-1}\bigl(r_1r_2\cdots r_{n-1} + r_1r_2\cdots r_{n-2}r_n +\cdots + r_2\cdots r_n\bigr).$$

*The constant coefficient is $$(-1)^n\bigl(r_1\cdots r_n\bigr).$$


But for two polynomials to be equal they have to be equal coefficient by coefficient. So that means that:
$$\begin{align*}
p_1 &= (-1)^1\bigl( r_1+\cdots + r_n\bigr)\\
p_2 &= (-1)^2\bigl( r_1r_2+r_1r_3+\cdots + r_1r_n + r_2r_3+\cdots + r_{n-1}r_n\bigr)\\
p_3 &= (-1)^3\bigl( r_1r_2r_3 + r_1r_2r_4+\cdots + r_1r_2r_n + r_1r_3r_4+\cdots + r_{n-2}r_{n-1}r_n\bigr)\\
&\vdots\\
p_n &= (-1)^n\bigl(r_1\cdots r_n\bigr).
\end{align*}$$
That is: 


*

*The sum of the roots is $-p_1$;

*The sum of all products of two roots is $(-1)^2p_2$;

*The sum of all products of three roots is $(-1)^3p_3$;

*The sum of all products of four roots is $(-1)^4p_4$;

*$\vdots$

*The sum of all products of $n-1$ roots is $(-1)^{n-1}p_{n-1}$;

*The product of all roots is $(-1)^np_n$.


Which is what the text you cite says.
How about a polynomial that is not monic,
$$p(x) = p_0x^n + p_1x^{n-1}+\cdots + p_{n-1}x + p_n,\qquad p_0\neq 0\ ?$$
Note that $r$ is a root of $p(x)$ if and only if it is a root of
$$P(x) = \frac{1}{p_0}p(x) = x^n + \frac{p_1}{p_0}x^{n-1} + \cdots + \frac{p_{n-1}}{p_0}x + \frac{p_n}{p_0}.$$
So the argument above applies to $P(x)$. 
