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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.