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For quantities $x_1,\dots ,x_n$, the $k$th elementary symmetric sum is $e_k={\sum }_{i_1<\cdots <i_k}{x}_{i_1}\cdots x_{i_k}$, with $e_0=1$. Every term uses $k$ distinct indices, and permuting the inputs leaves the sum unchanged.

For three inputs, $e_1=x_1+x_2+x_3$, $e_2=x_1{x}_2+x_1{x}_3+x_2{x}_3$, and $e_3=x_1{x}_2{x}_3$. Expanding the product gives ${\prod }_i(t-x_i)={\sum }_{k=0}^n(-1{)}^k{e}_k{t}^{n-k}$. These are Vieta's formulas, linking coefficients of a polynomial to symmetric expressions in its roots.

For a purely imaginary candidate $z=it$, $t\in \mathbb{R}$, the imaginary part of the polynomial is $2t(t^2-4)$. The candidate $t=0$ is not a root, while $t=\pm 2$ both make the real part zero. The resulting factorization into quadratic polynomials givesThe roots of a polynomial sum to $0+2=2$ and have product $4\cdot 4=16$, as required by Vieta's formulas for a monic quartic with cubic coefficient $-2$ and constant coefficient $16$.

For a monic polynomial of degree five, Vieta's formulas give the product of its polynomial roots as $(-1{)}^5$ times its constant coefficient. Therefore

By Vieta's formulas, the sum of all five polynomial roots, counted with multiplicity of a root, is minus the coefficient of $z^4$ divided by the leading coefficient. Hence

The two real slope roots of $p^2+fp-1=0$ are distinct because their discriminant is $f^2+4>0$. Vieta's formulas give $p_{+}{p}_{-}=-1$, so the tangent vectors $(1,p_{+})$ and $(1,p_{-})$ have zero scalar product. Whenever the coefficients define local solution curves, the two families are therefore orthogonal trajectories.

For the stated example at $(3,4)$, the slopes are $3$ and $-1/3$. In the family from part a, the constants are $1$ and $-1/9$, givingBoth curves pass through $(3,4)$, and $3(-1/3)=-1$ verifies the right angle.

By Vieta's formulas, the elementary symmetric polynomials in the roots giveFor each root, ${\alpha }_j^3+c_1{\alpha }_j^2+c_2{\alpha }_j+c_3=0$. Multiplying by ${\alpha }_j^k$ and summing gives the power sum recurrenceIts initial values are $s_0=3$, $s_1=-c_1$ and $s_2=c_1^2-2c_2$. They are polynomials with integer coefficients in $c_1,c_2,c_3$. The recurrence uses only addition and multiplication by those coefficients, so induction proves the same for every $s_k$. This is the cubic case of Newton's identities.

For $c_1=0$, put ${\alpha }_1=u$, ${\alpha }_2=v$, ${\alpha }_3=-u-v$. Then $c_2=-(u^2+uv+v^2)$ and $c_3=uv(u+v)$. Expanding the product of root differences givesThis is the discriminant of a depressed cubic in depressed form, and the displayed identity also proves its expression in integer-coefficient symmetric polynomials.

The Fundamental theorem of symmetric polynomials states that every symmetric polynomial in $n$ variables over a field is a polynomial in the elementary symmetric polynomials, uniquely so. For the three roots, Vieta's formulas give $e_1=0$, $e_2=p$, $e_3=-q$.

The squared product of root differences is symmetric. Expansion in elementary symmetric polynomials givesOne can verify the displayed identity simply by substituting the definitions of $e_1,e_2,e_3$ and expanding; it is the cubic polynomial discriminant.

If the cubic is irreducible, its roots are distinct in characteristic zero, so $\delta \ne 0$. Its Galois group is a transitive subgroup of $S_3$, hence either $A_3$ or $S_3$. A permutation of the roots multiplies $\delta$ by its sign. The fixed-field assertion of the Fundamental theorem of Galois theory therefore gives $\delta \in \mathbb{Q}$ exactly when all group elements are even. ThusFor example, $f=x^3-3x-1$ has no rational root: neither $1$ nor $-1$ is a root, and these are the only possibilities by the rational root theorem. Its discriminant is $108-27=81$, so $\delta =\pm 9\in \mathbb{Q}$ and its group is $A_3$.

The intended hypothesis must be a proper monic factor, with $1\le r\le p-1$. Without that restriction the printed deduction is false: every polynomial has the displayed trivial factorization with $r=p$, even when it has no root in $F$.

By part 1, the roots in a splitting field are distinct and have the form $\alpha +j$. The degree-$r$ monic factor is the product of $r$ corresponding linear factors. Comparing the coefficient of $X^{r-1}$ using Vieta's formulas givesBoth $a_1$ and the sum of prime-field elements lie in $F$. Since $1\le r<p$, the scalar $r$ is nonzero and invertible in $F$. Therefore $\alpha =(-a_1-{\sum }_{j\in J}j)/r\in F$, and part 1 proves the polynomial splits into distinct linear factors over $F$. In particular an Artin–Schreier polynomial of prime degree is either irreducible or splits completely.

Vieta's formulas connect two descriptions of a polynomial: its coefficients and its roots of a polynomial. They let us read the sum and product of the roots of a polynomial without finding the individual roots of a polynomial, and turn a proposed factorization of polynomials into checkable coefficient identities. The key is that coefficients encode elementary symmetric polynomials, which are unchanged by a permutation of the roots of a polynomial.

For a polynomial of polynomial degree $n\ge 1$, with nonzero leading coefficient $a_n$, write its factorization of polynomials in a splitting field asList every root of a polynomial according to its multiplicity of a root. ThenIn particular, ${\sum }_i{r}_i=-a_{n-1}/a_n$ and ${\prod }_i{r}_i=(-1{)}^n{a}_0/a_n$. With coefficients in the real numbers or complex numbers, the fundamental theorem of algebra ensures that a list of $n$ roots of a polynomial in the complex numbers exists when multiplicity of a root is counted.^[ref]

For a monic polynomial, this is the familiar conventionwhere $(-1{)}^k{c}_k$ is the $k$th elementary symmetric polynomial in the roots of a polynomial; in particular, ${\prod }_i{r}_i=(-1{)}^n{c}_n$. The ratio by $a_n$ in the general statement matters: multiplying every coefficient by the same nonzero scalar preserves both the roots of a polynomial and these ratios.

A short derivation explains every alternating sign. In the product ${\prod }_i(x-r_i)$, a term of polynomial degree $n-k$ arises by choosing $-r_i$ from $k$ distinct factor positions and $x$ from the others. Summing the resulting products gives $(-1{)}^k{e}_k$. Values at different positions may coincide: the selection is of positions, not distinct root values. Matching coefficients gives the formula. This is the distributive property organized by the number of chosen root factors.