How to prove relationship between coefficients and roots of a cubic (or) general polynomial

Question

I know about quadratic polynomial that $$\alpha +\beta =-\frac{b}{a}$$where $\alpha ,\beta $ are roots of the polynomial $ax^2+bx+c$

I also know it about cubic polynomial that $$\alpha+\beta+\gamma = -\frac{b}{a}$$where $\alpha,\beta,\gamma$ are roots of the polynomial $ax^3+bx^2+cx+d$

I can prove it for quadratic polynomial because I know how to calculate roots of the polynomial in terms of $a,b,c$, by quadratic formula

But I can't do it for cubic polynomial because I am afraid of the horror "cubic formula"

Also it will become more terrible when we talk about general polynomial of $n$ degree

I googled for proof of it but didn't got sufficient results

So please help me to prove the relationship between coefficients and roots of the cubic polynomial and further for general polynomial.

Answer 1

It's much easier to work from the roots to the coefficients than the other way around.

If you have a cubic polynomial $ax^3+bx^2+cx+d$ with roots $\alpha$, $\beta$, $\gamma$, then consider $$ a(x-\alpha)(x-\beta)(x-\gamma) $$ If you multiply this out you get a cubic polynomial that also has $\alpha$, $\beta$, $\gamma$ as roots and has $a$ as the leading coefficient. And this polynomial has to be the same polynomial as the one you started with, because otherwise subtracting them would produce a nonzero polynomial or degree $2$ or less that has $3$ roots, which is impossible.

Thus, multiplying out the above formula will give you all the coefficients as functions of the roots. Dividing through with $a$ then produces, among other things, the formula you're after. And this works for any degree.

(If there are multiple roots, then the above argument doesn't work 100% -- but in order to fix it you'd need to have a better definition for what it means in the first place for the roots to be, say, 2,2,3 instead of 2,3,3).

Answer 2

Remember that a polynomial can be written in two forms; one as a sum or as a product form with the zeroes of the polynomial as in some polynomial P could be

$$P(x) = \sum_{i = 0}^n a_ix^{i} = a_n\prod_{i=1}^n(x - x_n)$$

for a nth degree polynomial where $x_n$ is it's nth root;

Expanding it gives the Vieta's formulae taking k coefficients' products at time to obtain the result. It's pretty straightforward I hope.