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:


We can set this equal to original polynomial:


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=$(-1)^n*(\frac{r_1}{a_n})*r_2*r_3*\cdots*r_n$


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.

  • 2
    $\begingroup$ $a_n(x-\frac{r_1}{an})$ should be $a_n(x-r_1)$ $\endgroup$ Jun 28, 2020 at 20:06
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    $\begingroup$ Cf. Vieta's formulas $\endgroup$ Jun 28, 2020 at 20:13
  • $\begingroup$ @J.W.Tanner Oh thanks a lot! This is exactly what I am looking for. $\endgroup$
    – Aiden Chow
    Jun 28, 2020 at 20:15

1 Answer 1


$a_n$ 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.

Then polynomial can be factored as:


It should read this.

Then polynomial can be factored as:


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


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