Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For a polynomial of degree n,say $$F(x)=a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3+....+a_{n}x^n$$

The sum of roots is $-\frac{a_{n-1}}{a_n}$ and the product of roots is $(-1)^n \frac{a_{0}}{a_n}$ where the $a_i%$ are the coefficients of the polynomial. I've been using these formulas for some problems, but I don't get why they are true. How are they derived? Thanks.

share|cite|improve this question
It's about factorization and polynomial division: if $r_1, \ldots, r_n$ are the roots, then $a_nx^n + \cdots + a_0 = a_n(x - r_1) \cdots (x - r_n)$. – Dylan Moreland May 19 '12 at 21:37
By the way, these are known as Vieta's formulae. – Rahul May 19 '12 at 21:42
up vote 5 down vote accepted

Let $p(x)$ be a polynomial of degree $n$ and $r_1,r_2,\ldots,r_n$ be the $n$ roots of $p(x)$. We have $$p(x)=a_n(x-r_1)(x-r_2)\cdots(x-r_n)$$ because if $p(c)=0$ then $(x-c)$ divides $p(x)$. Multiplying this out, we see that the coefficient of $x^{n-1}$ (that is to say, $a_{n-1}$) is $-a_n\sum\limits_{i=1}^n r_i$ and the constant term ($a_0$) is $a_n(-r_1)(-r_2)\cdots (-r_n)=(-1)^na_nr_1r_2\cdots r_n$, which gives us the formula $$\sum\limits_{i=1}^n r_i = -\frac{a_{n-1}}{a_n} \text{ and }r_1r_2\cdots r_n=(-1)^n\frac{a_0}{a_n}.$$

share|cite|improve this answer
It's worth emphasis that this depends crucially on the fact that the coefficient ring is a domain, e.g. $\rm\:mod\ 8\!:\ x^2 - 1\:$ has roots $\rm\:x = 1,3\:$ but $\rm\:x^2-1\neq (x-1)(x-3).\ \ $ – Bill Dubuque May 19 '12 at 22:03
@BillDubuque And that it requires commutativity as well. But I didn't bother mentioning that in an answer to an "algebra-precalculus" question. – Alex Becker May 20 '12 at 4:44
Nowadays algebra-precalculus may include some modular arithmetic (in fact it did even long ago when I was a student). – Bill Dubuque May 20 '12 at 7:07
@BillDubuque I suppose so, at least pre-calc at some universities does (although not at my high school). But I've helped a lot of people with precalc homework from various classes at various institutions and have never seen them work in the polynomial ring $\mathbb Z_n[x]$. – Alex Becker May 20 '12 at 7:15

We can write the polynomial as $\,p(x)=a_0+a_1x+...+a_nx^n=a_n(x-\alpha_1)\cdot ...\cdot (x-\alpha_n)\,$ , with $\,\alpha_i\,$ the polynomial's root in some extension of the base field, thus it is just a matter of comparing coefficients in both sides...

share|cite|improve this answer

$P(x)=a_nx^n+a_{n-1}x^{n-1}+a_1x+a_0=a_n(x-x_1)(x-x_2)\cdot \cdot (x-x_n)$

To show the product of roots:

Need to calculate $P(0)$

$P(0)=a_n0^n+a_{n-1}0^{n-1}+a_10+a_0=a_0$ $P(0)=a_n(0-x_1)(0-x_2)\cdot \cdot (0-x_n)=a_n(-x_1)(-x_2)\cdot \cdot (-x_n)=a_n(-1)x_1(-1)x_2\cdot \cdot (-1)x_n=a_n(-1)^nx_1x_2\cdot \cdot x_n$

$P(0)=a_n(-1)^nx_1x_2\cdot \cdot x_n=a_0$

$x_1x_2\cdot \cdot x_n=\frac{a_0(-1)^n}{a_n}$

To show Sum of roots: Need to focus the coefficient of $x^{n-1}$


$(x-x_1)(x-x_2)(x-x_3)=(x^2-(x_1+x_2)x+x_1x_2)(x-x_3)=x^3-(x_1+x_2)x^2-x_3x^2 \cdot \cdot=x^3-(x_1+x_2+x_3)x^2 \cdot \cdot$

$(x-x_1)(x-x_2)(x-x_3)(x-x_4)=(x^3-(x_1+x_2+x_3)x^2 \cdot \cdot)(x-x_4)=x^4-(x_1+x_2+x_3)x^3-x_4x^3 \cdot \cdot=x^4-(x_1+x_2+x_3+x_4)x^3 \cdot \cdot\cdot \cdot$

$a_n(x-x_1)(x-x_2)\cdot \cdot (x-x_n)=a_n(x^n-(x_1+x_2+x_3+x_4+\cdot \cdot+x_n)x^{n-1} \cdot \cdot \cdot \cdot )=a_nx^n-a_n(x_1+x_2+x_3+x_4+\cdot \cdot+x_n)x^{n-1} \cdot \cdot \cdot \cdot=a_nx^n+a_{n-1}x^{n-1}+a_1x+a_0$

If we equal the coefficient of $x^{n-1}$ the result is: $-a_n(x_1+x_2+x_3+x_4+\cdot \cdot+x_n)=a_{n-1}$

$x_1+x_2+x_3+x_4+\cdot \cdot+x_n=\frac{-a_{n-1}}{a_n}$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.