# §1.11 Zeros of Polynomials

## §1.11(i) Division Algorithm

### Horner’s Scheme

Let

 1.11.1 $f(z)=a_{n}z^{n}+a_{n-1}z^{n-1}+\dots+a_{0}.$ Symbols: $z$: variable and $n$: nonnegative integer Referenced by: §1.11(i) Permalink: http://dlmf.nist.gov/1.11.E1 Encodings: TeX, pMML, png

Then

 1.11.2 $f(z)=(z-\alpha)(b_{n}z^{n-1}+b_{n-1}z^{n-2}+\dots+b_{1})+b_{0},$ Symbols: $z$: variable and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.11.E2 Encodings: TeX, pMML, png

where $b_{n}=a_{n}$,

 1.11.3 $b_{k}=\alpha b_{k+1}+a_{k},$ $k=n-1,n-2,\dots,0$, Symbols: $k$: integer and $n$: nonnegative integer Referenced by: §1.11(i) Permalink: http://dlmf.nist.gov/1.11.E3 Encodings: TeX, pMML, png
 1.11.4 $f(\alpha)=b_{0}.$ Permalink: http://dlmf.nist.gov/1.11.E4 Encodings: TeX, pMML, png

### Extended Horner Scheme

With $b_{k}$ as in (1.11.1)–(1.11.3) let $c_{n}=a_{n}$ and

 1.11.5 $c_{k}=\alpha c_{k+1}+b_{k},$ $k=n-1,n-2,\dots,1$. Symbols: $k$: integer and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.11.E5 Encodings: TeX, pMML, png

Then

 1.11.6 $f^{\prime}(\alpha)=c_{1}.$ Permalink: http://dlmf.nist.gov/1.11.E6 Encodings: TeX, pMML, png

More generally, for polynomials $f(z)$ and $g(z)$, there are polynomials $q(z)$ and $r(z)$, found by equating coefficients, such that

 1.11.7 $f(z)=g(z)q(z)+r(z),$ Symbols: $z$: variable, $q(z)$: polynomial and $r(z)$: polynomial Referenced by: §1.11(i) Permalink: http://dlmf.nist.gov/1.11.E7 Encodings: TeX, pMML, png

where $0\leq\deg r(z)<\deg g(z)$.

## §1.11(ii) Elementary Properties

A polynomial of degree $n$ with real or complex coefficients has exactly $n$ real or complex zeros counting multiplicity. Every monic (coefficient of highest power is one) polynomial of odd degree with real coefficients has at least one real zero with sign opposite to that of the constant term. A monic polynomial of even degree with real coefficients has at least two zeros of opposite signs when the constant term is negative.

### Descartes’ Rule of Signs

The number of positive zeros of a polynomial with real coefficients cannot exceed the number of times the coefficients change sign, and the two numbers have same parity. A similar relation holds for the changes in sign of the coefficients of $f(-z)$, and hence for the number of negative zeros of $f(z)$.

### Example

 1.11.8 $\displaystyle f(z)$ $\displaystyle=z^{8}+10z^{3}+z-4,$ $\displaystyle f(-z)$ $\displaystyle=z^{8}-10z^{3}-z-4.$ Symbols: $z$: variable Permalink: http://dlmf.nist.gov/1.11.E8 Encodings: TeX, TeX, pMML, pMML, png, png

Both polynomials have one change of sign; hence for each polynomial there is one positive zero, one negative zero, and six complex zeros.

Next, let $f(z)=a_{n}z^{n}+a_{n-1}z^{n-1}+\dots+a_{0}$. The zeros of $z^{n}f(1/z)=a_{0}z^{n}+a_{1}z^{n-1}+\dots+a_{n}$ are reciprocals of the zeros of $f(z)$.

The discriminant of $f(z)$ is defined by

 1.11.9 $D=a_{n}^{2n-2}\prod_{j Defines: $D$: discriminant of $f(z)$ (locally) Symbols: $z$: variable, $j$: integer, $k$: integer and $n$: nonnegative integer Referenced by: §1.11(ii) Permalink: http://dlmf.nist.gov/1.11.E9 Encodings: TeX, pMML, png

where $z_{1},z_{2},\dots,z_{n}$ are the zeros of $f(z)$. The elementary symmetric functions of the zeros are (with $a_{n}\not=0$)

 1.11.10 $\displaystyle z_{1}+z_{2}+\dots+z_{n}$ $\displaystyle=-a_{n-1}/a_{n},$ $\displaystyle\sum_{1\leq j $\displaystyle=a_{n-2}/a_{n},$ $\displaystyle\mathrel{\vdots}$ $\displaystyle z_{1}z_{2}\cdots z_{n}$ $\displaystyle=(-1)^{n}a_{0}/a_{n}.$

## §1.11(iii) Polynomials of Degrees Two, Three, and Four

The roots of $az^{2}+bz+c=0$ are

 1.11.11 $\frac{-b\pm\sqrt{D}}{2a},$ $\displaystyle D$ $\displaystyle=b^{2}-4ac.$ Symbols: $D$: discriminant Permalink: http://dlmf.nist.gov/1.11.E11 Encodings: TeX, TeX, pMML, pMML, png, png

The sum and product of the roots are respectively $-b/a$ and $c/a$.

### Cubic Equations

Set $z=w-\tfrac{1}{3}a$ to reduce $f(z)=z^{3}+az^{2}+bz+c$ to $g(w)=w^{3}+pw+q$, with $p=(3b-a^{2})/3$, $q=(2a^{3}-9ab+27c)/27$. The discriminant of $g(w)$ is

 1.11.12 $D=-4p^{3}-27q^{2}.$ Defines: $D$: discriminant (locally) Symbols: $p$ and $q$ A&S Ref: 3.8.1 Permalink: http://dlmf.nist.gov/1.11.E12 Encodings: TeX, pMML, png

Let

 1.11.13 $\displaystyle A$ $\displaystyle=\sqrt[3]{-\tfrac{27}{2}q+\tfrac{3}{2}\sqrt{-3D}},$ $\displaystyle B$ $\displaystyle=-3p/A.$ Defines: $A$ (locally) and $B$ (locally) Symbols: $p$, $q$ and $D$: discriminant Permalink: http://dlmf.nist.gov/1.11.E13 Encodings: TeX, TeX, pMML, pMML, png, png

The roots of $g(w)=0$ are

 1.11.14 $\tfrac{1}{3}(A+B),$ $\tfrac{1}{3}(\rho A+\rho^{2}B),$ $\tfrac{1}{3}(\rho^{2}A+\rho B),$ Symbols: $\rho$, $A$ and $B$ Permalink: http://dlmf.nist.gov/1.11.E14 Encodings: TeX, TeX, TeX, pMML, pMML, pMML, png, png, png

with

 1.11.15 $\displaystyle\rho$ $\displaystyle=-\tfrac{1}{2}+\tfrac{1}{2}\sqrt{-3}=e^{2\pi i/3},$ $\displaystyle\rho^{2}$ $\displaystyle=e^{-2\pi i/3}.$ Symbols: $e$: base of exponential function and $\rho$ Permalink: http://dlmf.nist.gov/1.11.E15 Encodings: TeX, TeX, pMML, pMML, png, png

Addition of $-\frac{1}{3}a$ to each of these roots gives the roots of $f(z)=0$.

### Example

$f(z)=z^{3}-6z^{2}+6z-2$, $g(w)=w^{3}-6w-6$, $A=3\sqrt[3]{4}$, $B=3\sqrt[3]{2}$. Roots of $f(z)=0$ are $2+\sqrt[3]{4}+\sqrt[3]{2}$, $2+\sqrt[3]{4}\rho+\sqrt[3]{2}\rho^{2}$, $2+\sqrt[3]{4}\rho^{2}+\sqrt[3]{2}\rho$.

For another method see §4.43.

### Quartic Equations

Set $z=w-\tfrac{1}{4}a$ to reduce $f(z)=z^{4}+az^{3}+bz^{2}+cz+d$ to

 1.11.16 $\displaystyle g(w)$ $\displaystyle=w^{4}+pw^{2}+qw+r,$ $\displaystyle p$ $\displaystyle=(-3a^{2}+8b)/8,$ $\displaystyle q$ $\displaystyle=(a^{3}-4ab+8c)/8,$ $\displaystyle r$ $\displaystyle=(-3a^{4}+16a^{2}b-64ac+256d)/256.$ Symbols: $w$: variable, $p$, $q$ and $r$ Permalink: http://dlmf.nist.gov/1.11.E16 Encodings: TeX, TeX, TeX, TeX, pMML, pMML, pMML, pMML, png, png, png, png

The discriminant of $g(w)$ is

 1.11.17 $D=16p^{4}r-4p^{3}q^{2}-128p^{2}r^{2}+144pq^{2}r-27q^{4}+256r^{3}.$ Defines: $D$: discriminant (locally) Symbols: $p$, $q$ and $r$ Permalink: http://dlmf.nist.gov/1.11.E17 Encodings: TeX, pMML, png

For the roots $\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}$ of $g(w)=0$ and the roots $\theta_{1},\theta_{2},\theta_{3}$ of the resolvent cubic equation

 1.11.18 $z^{3}-2pz^{2}+(p^{2}-4r)z+q^{2}=0,$

we have

 1.11.19 $\displaystyle 2\alpha_{1}$ $\displaystyle=\phantom{+}\sqrt{-\theta_{1}}+\sqrt{-\theta_{2}}+\sqrt{-\theta_{% 3}},$ $\displaystyle 2\alpha_{2}$ $\displaystyle=\phantom{+}\sqrt{-\theta_{1}}-\sqrt{-\theta_{2}}-\sqrt{-\theta_{% 3}},$ $\displaystyle 2\alpha_{3}$ $\displaystyle=-\sqrt{-\theta_{1}}+\sqrt{-\theta_{2}}-\sqrt{-\theta_{3}},$ $\displaystyle 2\alpha_{4}$ $\displaystyle=-\sqrt{-\theta_{1}}-\sqrt{-\theta_{2}}+\sqrt{-\theta_{3}}.$ Defines: $\theta_{j}$: cubic roots (locally) Symbols: $j$: integer Permalink: http://dlmf.nist.gov/1.11.E19 Encodings: TeX, TeX, TeX, TeX, pMML, pMML, pMML, pMML, png, png, png, png

The square roots are chosen so that

 1.11.20 $\sqrt{-\theta_{1}}\;\sqrt{-\theta_{2}}\;\sqrt{-\theta_{3}}=-q.$ Symbols: $q$ and $\theta_{j}$: cubic roots Permalink: http://dlmf.nist.gov/1.11.E20 Encodings: TeX, pMML, png

Add $-\tfrac{1}{4}a$ to the roots of $g(w)=0$ to get those of $f(z)=0$.

### Example

$f(z)=z^{4}-4z^{3}+5z+2$,   $g(w)=w^{4}-6w^{2}-3w+4$. Resolvent cubic is $z^{3}+12z^{2}+20z+9=0$ with roots $\theta_{1}=-1$, $\theta_{2}=-\tfrac{1}{2}(11+\sqrt{85})$, $\theta_{3}=-\tfrac{1}{2}(11-\sqrt{85})$, and $\sqrt{-\theta_{1}}=1$, $\sqrt{-\theta_{2}}=\tfrac{1}{2}(\sqrt{17}+\sqrt{5})$, $\sqrt{-\theta_{3}}=\tfrac{1}{2}(\sqrt{17}-\sqrt{5})$. So $2\alpha_{1}=1+\sqrt{17}$, $2\alpha_{2}=1-\sqrt{17}$, $2\alpha_{3}=-1+\sqrt{5}$, $2\alpha_{4}=-1-\sqrt{5}$, and the roots of $f(z)=0$ are $\tfrac{1}{2}(3\pm\sqrt{17})$, $\tfrac{1}{2}(1\pm\sqrt{5})$.

## §1.11(iv) Roots of Unity and of Other Constants

The roots of

 1.11.21 $z^{n}-1=(z-1)(z^{n-1}+z^{n-2}+\dots+z+1)=0$ Symbols: $z$: variable and $n$: nonnegative integer Referenced by: §1.3(ii) Permalink: http://dlmf.nist.gov/1.11.E21 Encodings: TeX, pMML, png

are $1$, $e^{2\pi i/n}$, $e^{4\pi i/n},\dots,e^{(2n-2)\pi i/n}$, and of $z^{n}+1=0$ they are $e^{\pi i/n},e^{3\pi i/n},\dots,e^{(2n-1)\pi i/n}$.

The roots of

 1.11.22 $z^{n}=a+ib,$ $a,b$ real, Symbols: $z$: variable and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.11.E22 Encodings: TeX, pMML, png

are

 1.11.23 $\sqrt[n]{R}\left(\mathop{\cos\/}\nolimits\!\left(\frac{\alpha+2k\pi}{n}\right)% +i\mathop{\sin\/}\nolimits\!\left(\frac{\alpha+2k\pi}{n}\right)\right),$

where $R=(a^{2}+b^{2})^{1/2}$, $\alpha=\mathop{\mathrm{ph}\/}\nolimits\!\left(a+ib\right)$, with the principal value of phase (§1.9(i)), and $k=0,1,\dots,n-1$.

## §1.11(v) Stable Polynomials

 1.11.24 $f(z)=a_{0}+a_{1}z+\dots+a_{n}z^{n},$ Symbols: $z$: variable and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.11.E24 Encodings: TeX, pMML, png

with real coefficients, is called stable if the real parts of all the zeros are strictly negative.

### Hurwitz Criterion

Let

 1.11.25 $\displaystyle D_{1}$ $\displaystyle=a_{1},$ $\displaystyle D_{2}$ $\displaystyle=\begin{vmatrix}a_{1}&a_{3}\\ a_{0}&a_{2}\end{vmatrix},$ $\displaystyle D_{3}$ $\displaystyle=\begin{vmatrix}a_{1}&a_{3}&a_{5}\\ a_{0}&a_{2}&a_{4}\\ 0&a_{1}&a_{3}\end{vmatrix},$ Symbols: $\det$: determinant and $D_{j}$: quantities Permalink: http://dlmf.nist.gov/1.11.E25 Encodings: TeX, TeX, TeX, pMML, pMML, pMML, png, png, png

and

 1.11.26 $D_{k}=\det[h_{k}^{(1)},h_{k}^{(3)},\dots,h_{k}^{(2k-1)}],$

where the column vector $h_{k}^{(m)}$ consists of the first $k$ members of the sequence $a_{m},a_{m-1},a_{m-2},\dots$ with $a_{j}=0$ if $j<0$ or $j>n$.

Then $f(z)$, with $a_{n}\not=0$, is stable iff $a_{0}\not=0$; $D_{2k}>0$, $k=1,\dots,\left\lfloor\frac{1}{2}n\right\rfloor$; $\mathop{\mathrm{sign}\/}\nolimits D_{2k+1}=\mathop{\mathrm{sign}\/}\nolimits a% _{0}$, $k=0,1,\dots,\left\lfloor\frac{1}{2}n-\frac{1}{2}\right\rfloor$.