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For the quadratic polynomial $p(x)=2x^2-6x+5$, the discriminant is $(-6{)}^2-4(2)(5)=-4$, so its roots of a polynomial are the complex conjugatesTheir sum is $3=-(-6)/2$, and their product is $(9+1)/4=5/2$, matching the constant coefficient divided by the leading coefficient. More revealingly, $r_{+}{r}_{-}=r_{+}\overline{r_{+}}=|r_{+}{|}^2$: the product is the square of a complex modulus. Coefficients in the real numbers therefore need not imply polynomial roots in the real numbers.

For any polynomial with coefficients in the real numbers, $p(\overline{z})=\overline{p(z)}$. Thus a nonreal root of a polynomial brings its complex conjugate along with the same multiplicity of a root. This follows by applying complex conjugation to the polynomial expression and its factors. It explains why sums of complex conjugate terms can produce coefficients in the real numbers.

A cubic equation is a polynomial equation of degree of a polynomial three, $a{x}^3+b{x}^2+cx+d=0$, with $a\ne 0$. Its three complex roots of a polynomial, counted with multiplicity of a root, need not all have physical significance in an application; a model often selects a branch by continuity from a known limit.

A cubic polynomial in one variable has polynomial degree three. Over a splitting field, its factorization of polynomials contains three polynomial factors of polynomial degree one when multiplicity of a root is counted.

For two polynomials over the complex numbers $p,q$ of equal positive degree of a polynomial $n$ with no common root of a polynomial, there are unique polynomials $u,v$ of degree of a polynomial less than $n$ such that $up+vq=1$. To prove the bound, let $V_m$ be the vector space of polynomials of degree of a polynomial at most $m$, including the zero polynomial. The linear map $(a,b)\mapsto ap+bq$ from $V_{n-1}\times V_{n-1}$ to $V_{2n-1}$ has zero kernel of a linear map: if $ap=-bq$, each root of a polynomial of $p$, counted with its multiplicity of a root, must occur in $b$ because it does not occur in $q$, forcing $b=0$ by the degree of a polynomial bound, and then $a=0$. Both vector spaces have dimension $2n$, so the rank-nullity theorem makes the linear map a bijection, and $1$ has a unique preimage.

Let $V(p)$ be the number of sign changes in the list of nonzero coefficients of a real polynomial, arranged by increasing or decreasing degree. The positive roots of a polynomial, counted with multiplicity of a root, satisfyThe bound follows by induction on the nonzero terms after removing the lowest power of $x$. Differentiation removes the constant coefficient. If it and the next coefficient have opposite signs, the Rolle root count with multiplicities loses at most one root and the sign-change count loses one. If their signs agree, the derivative must have an extra root before the first positive root, so neither count needs that extra allowance. The parity statement follows by comparing the signs near zero and at positive infinity: roots of odd multiplicity reverse the sign, and roots of even multiplicity preserve it. Apply the same rule to $p(-x)$ to bound negative roots.

A root of a polynomial is double when its multiplicity of a root is exactly two. Equivalently, $p(x)=(x-x_0{)}^{2}q(x)$ with $q(x_0)\ne 0$. A generic small perturbation splits it according to square-root splitting of a double polynomial root. Over a field of characteristic different from two, the derivative test is $p(a)=p^{\prime }(a)=0$ and $p^{\prime \prime }(a)\ne 0$. In characteristic two, use the Hasse derivative: a double root of $(X-a{)}^2$ has zero second ordinary formal derivative but nonzero second Hasse derivative.

If $P=z^{d}p$ and $p$ is nonnegative on the unit circle, a root of a polynomial $\zeta =e^{i{\theta }_0}$ there has even multiplicity of a root. Indeed $p(e^{i\theta })$ is a real analytic nonnegative function, whose first nonzero term in its Taylor series at a zero must have even order. The map $\theta \mapsto e^{i\theta }-\zeta$ has a simple zero, and multiplication by $e^{-id\theta }$ is nonvanishing, so the order equals the polynomial multiplicity.

Two real polynomials with only real roots of a polynomial interlace when their ordered roots alternate, counting multiplicity of a root. For equal polynomial degrees, one ordering is the displayed chain; the opposite ordering is also permitted. Consecutive degrees give the analogous chain with one extra root at each end for the higher-degree polynomial. Strict interlacing uses strict inequalities. Weak interlacing permits common roots and follows as a limit of strict interlacing.

Three explicit functions separate the hypotheses from the conclusion. These are original mathematical illustrations, not measured data.

The polynomial function $h(x)=x^3-x=x(x-1)(x+1)$ is a continuous function on $[-3/2,3/2]$, with $h(-3/2)=-15/8$ and $h(3/2)=15/8$. The intermediate value theorem guarantees at least one zero of a function in the interior, while the polynomial factorization displays three: $-1,0,1$. Thus strict opposite signs do not establish uniqueness.

The everywhere-defined functionhas $j(-1)=-1$ and $j(1)=1$, yet no zero of a function. Its jump discontinuity at zero is exactly where the continuity hypothesis fails. A drawn vertical connector would be misleading: the function graph contains neither the intermediate heights nor a zero of a function at that jump. Even continuity only on the interior is insufficient: setting $q(x)=-1$ for $0\le x<1$ and $q(1)=1$ gives opposite endpoint signs on $[0,1]$ with no zero of a function, because the left one-sided limit at $1$ is not $q(1)$.

The polynomial function $t(x)=x^2$ on $[-1,1]$ is a continuous function, and $t(-1)=t(1)=1$, yet $t(0)=0$. Its root of a polynomial at zero has multiplicity of a root two. The function graph touches its horizontal tangent line there and stays positive on both sides: existence of a zero of a function does not require a sign change. More generally, near a real root of a polynomial $r$, a polynomial factorization $p(x)=(x-r{)}^{2m}u(x)$ with $m\ge 1$ and $u(r)\ne 0$ has the same sign on both sides sufficiently close to $r$, since the continuous function $u$ retains its sign. The intermediate value theorem does not rule out values outside the range between endpoint outputs.^[ref]

With $P=W^{\prime }$ a nonconstant polynomial, positive area measure and Berezin integration oriented positively, a holomorphic polynomial insertion in $S=|P{|}^2-P^{\prime }{\theta }_1{\theta }_2-\overline{P^{\prime }}{\bar{\theta }}_1{\bar{\theta }}_2$ equals $\pi {\sum }_a{m}_{a}f(z_a)$, where $P(z_a)=0$ and $m_a$ is the multiplicity of a root. Rescaling $P$ by $t>0$ leaves the insertion fixed and localizes it as $t\to \infty$.

A root is multiple when its multiplicity of a root is greater than one.

The coefficients are real, so the supplied polynomial root $2+2i$ has its complex conjugate $2-2i$ as another polynomial root. Their monic polynomial factor of degree two isPolynomial division then givesThus all five polynomial roots, with multiplicity of a root, areTheir sum is $5$ and their product is $8\cdot 1\cdot (-1)=-8$, agreeing with solution and solution.

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

Using the printed convention that the projective points represent lines of binary quartics, the locus of GIT stable points isThe root at infinity is included. We determine stability by calculating every relevant algebraic one-parameter subgroup weight.

Use the natural action $(g\cdot f)(X,Y)=f(g^{-1}(X,Y))$. Every nontrivial algebraic one-parameter subgroup of $S{L}_2$ is conjugate to ${\lambda }_m(t)=\mathrm{diag}(t^m,t^{-m})$ for some integer $m>0$: its two integral weights of a representation sum to zero, and a basis of weight vectors can be rescaled to have determinant one. WriteIf $r=\min \{j:a_j\ne 0\}$, then $r$ is the multiplicity of a root at the projective point $[1:0]$: in the local coordinate $Y/X$, the polynomial begins with a nonzero multiple of $(Y/X{)}^r$.

The Hilbert-Mumford criterion for projective stability, with $\mu =-\min$(active weights), now givesBeing a GIT stable point requires this number to be strictly positive for every nontrivial subgroup. Conjugating the subgroup lets $[1:0]$ range over the entire projective line. Thus the root-multiplicity criterion for stable binary forms becomes $r_p(f)<2$ at every point $p$. A nonzero degree-four form over $\mathbb{C}$ has four projective roots counted with multiplicity, so this means precisely four distinct roots. This is equivalently nonvanishing of its discriminant.

The boundary cases show why a weak inequality would give the wrong answer. A triple or quadruple root can be moved to $[1:0]$, leaving only strictly positive weights; its affine representative tends to zero under that subgroup, so it is unstable. If all root multiplicities are at most two, all Hilbert-Mumford values are nonnegative and the point is a GIT semistable point. A double root gives a zero value for a suitable subgroup, so never represents a GIT stable point. More concretely, after putting that double root at $[1:0]$ one hasand ${\lambda }_1(t)\cdot f\to a{X}^2{Y}^2$. When there are three distinct roots, this limiting form, representing a GIT semistable point with two distinct roots, is outside the original group orbit, so that group orbit is not closed in the semistable locus. When there are two double roots, a change of coordinates gives $X^2{Y}^2$, which has the positive-dimensional diagonal torus as stabilizer. Neither case represents a GIT stable point.

For a form with four distinct roots, the projective stabilizer is finite: it permutes the four roots, and a Möbius transformation fixing three distinct points is the identity. Thus its image in $PG{L}_2$ injects into the finite permutation group on the roots. The kernel of a group homomorphism for $S{L}_2\to PG{L}_2$ is $\{\pm I\}$, also finite. This agrees with the finite-stabilizer condition in geometric invariant theory. The weight calculation already supplies the required strict stability condition for every conjugate subgroup.

Rescale $W\mapsto tW$ with $t>0$. After Berezin integration, the unnormalized holomorphic function insertion isIts derivative is an ordinary total derivative:The last equality again follows from polynomial growth and exponential decay. Thus the insertion is independent of $t$, and supersymmetric localization allows evaluation as $t\to \infty$ at the zeros of $P$, as in localization of a zero-dimensional polynomial model.

For a simple zero $z_a$, $P(z)=P^{\prime }(z_a)(z-z_a)+O((z-z_a{)}^2)$. The local Gaussian integral cancels the factor $|P^{\prime }(z_a){|}^2$ and contributes $\pi f(z_a)$. For a zero with multiplicity of a root $m_a$, write $P(z)=a_a(z-z_a{)}^{m_a}+\cdots$. Its leading radial integral isThe insertion tends to $f(z_a)$ there; contributions away from the zeros vanish. Therefore the full answer, including degenerate critical points, isThis is an unnormalized expectation. For simple zeros all $m_a=1$; division by $Z=\pi r$ would give the normalized average, but is not part of the requested insertion. Taking $f=1$ recovers the partition function, and taking $f=Pg$ recovers its vanishing supersymmetric Ward identity.

Pair the off-circle roots of a polynomial using reciprocal-conjugate root pairing, and split each unit-circle root of a polynomial's even multiplicity equally between the two members of a pair. The fundamental theorem of algebra and the leading coefficient $p_d$ give $P(z)=p_d{\prod }_{i=1}^d(z-{\zeta }_i)(z-1/\overline{{\zeta }_i})$. None of the selected ${\zeta }_i$ is zero.

On the unit circle, the identityturns $P(z)/z^d$ intoAt a point of the unit circle outside the finite set of roots of a polynomial, the product is positive and $p(z)$ is nonzero and nonnegative. Its ratio to the product is therefore real and strictly positive. This proves $c>0$, even though the algebraic expression initially permits a complex constant. ConsequentlyThis proves the Fejér–Riesz theorem for a nonzero trigonometric polynomial of actual order $d$. A positive constant has a constant square-root factor, and the identically zero trigonometric polynomial has $q=0$; if $p_d=0$ for a specified upper order $d$, reduce to the actual order first.

Although the PDF permits assuming even multiplicity, there is a short proof of even multiplicity of unit-circle roots of a nonnegative trigonometric polynomial. The real analytic function $p(e^{i\theta })\ge 0$ cannot have a zero of odd order. Near $\zeta =e^{i{\theta }_0}$, $e^{i\theta }-\zeta$ has a simple zero, and the nonzero factor $e^{-id\theta }$ leaves the zero order of $P(e^{i\theta })$ unchanged. Hence the multiplicity of a root must be even.

For $z\ne 0$, the coefficient symmetry yieldsThus $P$ equals its reversed-conjugate polynomial. If $\zeta \ne 0$ is a root of a polynomial, with $P(\zeta )=0$, this identity gives $P(1/\overline{\zeta })=0$, provingThe reciprocal-conjugate root pairing also preserves the multiplicity of a root: reversal and complex conjugation take each factor associated with $\zeta$ to the corresponding factor associated with $1/\overline{\zeta }$, with the same exponent.

Since $P(0)=p_{-d}\ne 0$, no zero root of a polynomial occurs, so the reciprocal operation is always defined on the roots of a polynomial. Roots off the unit circle are paired on opposite sides of it; roots of a polynomial on the unit circle are fixed by this operation. Pairing alone does not force even multiplicity at those fixed roots of a polynomial; that extra fact uses nonnegativity.

Suppose, for a contradiction, that $p$ and $q$ have no common root of a polynomial. Since the polynomial ring $\mathbb{C}[x]$ has no zero divisors and $p,q\ne 0$, the relation $ap+bq=0$ with $a,b$ not both zero forces both $a$ and $b$ to be nonzero.

By the fundamental theorem of algebra, writewhere the ${\alpha }_j$ are distinct. For each $j$, $q({\alpha }_j)\ne 0$. The equality $bq=-ap$ implies that $(x-{\alpha }_j{)}^{m_j}$ divides $bq$. The multiplicity of a root in polynomial multiplication is the sum of its multiplicities in the factors: factor out the largest powers of $x-{\alpha }_j$ and note that the two remaining factors have nonzero values at ${\alpha }_j$. Since $q({\alpha }_j)\ne 0$, $b$ must contain $(x-{\alpha }_j{)}^{m_j}$.

Thus $b$ has at least ${\sum }_j{m}_j=n$ roots of a polynomial counted with multiplicity of a root, so $\mathrm{deg}b\ge n$. This contradicts $b\in V_{n-1}$. ConsequentlyCounting multiplicity of a root is necessary: counting only distinct roots of a polynomial would miss repeated factors.

Let $p(x)=x^3-5x-1$. Its values at useful points areSince a polynomial is a continuous function, the intermediate value theorem gives one polynomial root in each of the disjoint intervals $(-3,-2)$, $(-1,0)$ and $(2,3)$. A nonzero cubic polynomial has at most three distinct polynomial roots. Hence it hasThese three distinct polynomial roots already exhaust its degree of a polynomial, so each has multiplicity of a root one.

Yes: for $k=2$ the stated inequality guarantees zero-stability. Part (ii) puts every polynomial root in the closed unit disc. The remaining root condition for a multistep method is that every root on the unit circle has multiplicity of a root one.

If this failed for a quadratic polynomial, it would have a repeated root $\xi$ with $|\xi |=1$, andConsequently $|{\alpha }_1|=2|{\alpha }_2|$ and $|{\alpha }_0|=|{\alpha }_2|$, contradictingbecause ${\alpha }_2\ne 0$. Thus all roots on the unit circle are simple andThis conclusion does not assert consistency of a numerical method: part (i) rules out consistency of a numerical method whenever the coefficient inequality is strict. Zero-stability and consistency of a numerical method are separate requirements.