Let's say we had a $n$th degree polynomial equation $a_{n}x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_2x^2+a_1x+a_0=0$, with $a$ being real coefficient. What would the sum and product of its roots be(in terms of $a$)? I think I got the product one but not the sum.
For the product:
Let's say the polynomial's roots are $r_1,r_2,r_3,\ldots,r_n$.
Then polynomial can be factored as:
$a_n(x-\frac{r_1}{a_n})(x-r_2)(x-r_3)\ldots(x-r_n)$
We can set this equal to original polynomial:
$a_n(x-\frac{r_1}{a_n})(x-r_2)(x-r_3)\ldots(x-r_n)=a_{n}x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_2x^2+a_1x+a_0=0$
Compare constant terms:
$a_{n}x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_2x^2+a_1x+a_0$ constant term=$a_0$.
$a_n(x-\frac{r_1}{a_n})(x-r_2)(x-r_3)\ldots(x-r_n)$ constant term=$(-1)^n*(\frac{r_1}{a_n})*r_2*r_3*\cdots*r_n$
$a_0=(-1)^n*(\frac{r_1}{a_n})*r_2*r_3*\cdots*r_n$
Multiply $(-1)^na_n$ both sides:
$r_1*r_2*r_3*\cdots r_n=(-1)^na_0a_n$
Is this correct? Also, what can I do for the sum of the roots(I think we use the coefficients of $x^{n-1}$)?
EDIT: J.W. Tanner has noted in his comment that this is Vieta's Formulas which is exactly what I was looking for but couldn't find.
Best Answer
$a_n$ is not a root, it is the leading coefficient.
Imagine the polynomial $(x-2)(x-3) = x^2 - 5x + 6$.
The leading coefficient is $1$ but the roots are $2$ and $3$.
This part of your argument is incorrect.
It should read this.
From this representation it gets obvious that the free term is
$a_n(-1)^nr_1 r_2 ... r_n$
But on the other hand we know it is $a_0$.
So the product of the roots must be:
$$\frac{a_0}{(-1)^na_n} = (-1)^n \cdot \frac{a_0}{a_n} $$
For calculating the sum of the roots just compare the coefficient before $x^{n-1}$.
You will get that the sum of the roots equals $$-\frac{a_{n-1}}{a_n}$$
See also: Vieta's formulas