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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.

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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
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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 $$\sum\limits_{i=1}^n r_i = -\frac{a_{n-1}}{a_n} \text{ and }r_1r_2\cdots r_n=(-1)^n\frac{a_0}{a_n}.$$

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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
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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...

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

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 $x^{n-1}$

$(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2$

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

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

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

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

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

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