# Quartic function

(Redirected from Quartic equation)
Graph of a polynomial of degree 4, with 3 critical points.

In mathematics, a quartic function, is a function of the form

$f(x)=ax^4+bx^3+cx^2+dx+e,$

where a is nonzero, which is defined by a polynomial of degree four, called quartic polynomial.

Sometimes the term biquadratic is used instead of quartic, but, usually, biquadratic function refers to a quadratic function of a square (or, equivalently, to the function defined by a quartic polynomial without terms of odd degree), having the form

$f(x)=ax^4+cx^2+e.$

A quartic equation, or equation of the fourth degree, is an equation consisting in equating to zero a quartic polynomial, of the form

$ax^4+bx^3+cx^2+dx+e=0 ,$

where a ≠ 0.

The derivative of a quartic function is a cubic function.

Since a quartic function is defined by a polynomial of even degree, it has the same infinite limit when the argument goes to positive or negative infinity. If a is positive, then the function increases to positive infinity at both ends; and thus the function has a global minimum. Likewise, if a is negative, it decreases to negative infinity and has a global maximum. In both cases it may have, but not always, another local maximum and another local minimum.

The degree four (quartic case) is the highest degree such that every polynomial equation can be solved by radicals.

## History

Lodovico Ferrari is attributed with the discovery of the solution to the quartic in 1540, but since this solution, like all algebraic solutions of the quartic, requires the solution of a cubic to be found, it couldn't be published immediately.[1] The solution of the quartic was published together with that of the cubic by Ferrari's mentor Gerolamo Cardano in the book Ars Magna (1545).

The Soviet historian I. Y. Depman claimed that even earlier, in 1486, Spanish mathematician Valmes was burned at the stake for claiming to have solved the quartic equation.[2] Inquisitor General Tomás de Torquemada allegedly told Valmes that it was the will of God that such a solution be inaccessible to human understanding.[3] However Beckmann, who popularized this story of Depman in the west, said that it was unreliable and hinted that it may have been invented as Soviet antireligious propaganda.[4] Beckmann's version of this story has been widely copied in several books and internet sites, usually without his reservations and sometimes with fanciful embellishments. Several attempts to find corroborating evidence for this story, or even for the existence of Valmes, have failed.[5]

The proof that four is the highest degree of a general polynomial for which such solutions can be found was first given in the Abel–Ruffini theorem in 1824, proving that all attempts at solving the higher order polynomials would be futile. The notes left by Évariste Galois prior to dying in a duel in 1832 later led to an elegant complete theory of the roots of polynomials, of which this theorem was one result.[6]

## Applications

Polynomials of high degrees often appear in problems involving optimization, and sometimes these polynomials happen to be quartics, but this is a coincidence.

Quartics often arise in computer graphics, for example when computing the intersection of two conic sections. Another example is ray-tracing against quartic surfaces such as tori.

In computer-aided manufacturing, the torus is a common shape associated with the endmill cutter. To calculate its location relative to a triangulated surface, the position of a horizontal torus on the Z-axis must be found where it is tangent to a fixed line, and this requires the solution of a general quartic equation to be calculated. Over 10% of the computational time in a CAM system can be consumed simply calculating the solution to millions of quartic equations.

## Solving a quartic equation

### Nature of the roots

Given the general quartic equation

$ax^4 + bx^3 + cx^2 + dx + e = 0$

with real coefficients and $a \ne 0,$ the nature of its roots is mainly determined by the sign of its discriminant

\begin{align} \Delta\ =\ &256 a^3 e^3 - 192 a^2 b d e^2 - 128 a^2 c^2 e^2 + 144 a^2 c d^2 e - 27 a^2 d^4 \\ &+ 144 a b^2 c e^2 - 6 a b^2 d^2 e - 80 a b c^2 d e + 18 a b c d^3 + 16 a c^4 e \\ &- 4 a c^3 d^2 - 27 b^4 e^2 + 18 b^3 c d e - 4 b^3 d^3 - 4 b^2 c^3 e + b^2 c^2 d^2 \end{align}

This may be refined by considering the signs of three other polynomials:

$P = 8ac - 3b^2$

such that $\tfrac{P}{8a^2}$ is the second degree coefficient of the associated depressed quartic (see below);

$\Delta_0 = c^2 - 3bd + 12ae$

which is 0 if the quartic has a triple root; and

$D = 64 a^3 e - 16 a^2 c^2 + 16 a b^2 c - 16 a^2 bd - 3 b^4$

which is 0 if the quartic has two double roots.

The possible cases for the nature of the roots are as follows:[7]

• If $\Delta < 0$ then the equation has two real roots and two complex conjugate roots.
• If $\Delta > 0$ then the equation's four roots are either all real or all complex.
• If P < 0 and D < 0 then all four roots are real and distinct.
• If P > 0 or D > 0 then there are two pairs of complex conjugate roots.[8]
• If $\Delta = 0$ then either the polynomial has a multiple root, or it is the square of a quadratic polynomial. Here are the different cases that can occur:
• If P < 0 and D < 0 and $\scriptstyle \Delta_0$ ≠ 0, there is a real double root and two real simple roots.
• If (P > 0 and D ≠ 0) or D > 0, there is a real double root and two complex conjugate roots.
• If $\scriptstyle \Delta_0$ = 0 and D ≠ 0, there is a triple root and a simple root, all real.
• If D = 0, then:
• If P < 0, there are two real double roots.
• If P > 0, there are two complex conjugate double roots.
• If $\scriptstyle \Delta_0$ = 0, all four roots are equal to $-\tfrac{b}{4a}$

There are some cases that do not seem to be covered, but they can not occur. For example $\scriptstyle \Delta$ > 0, P = 0 and D ≤ 0 is not one of the cases. However if $\scriptstyle \Delta$ > 0 and P = 0 then D > 0 so this combination is not possible.

### General formula for roots

Quartic formula written out in full. This formula is too unwieldy for general use, hence other methods or simpler formulas are generally used.[9]

The four roots ($x_1, x_2, x_3, x_4$) for the general quartic equation

$ax^4+bx^3+cx^2+dx+e=0 \,$

with a ≠ 0 are given in the following formula, which is deduced from the one in the section Solving by factoring into quadratics by back changing the variables (see section Converting to a depressed quartic) and using the formulas for the quadratic and cubic equations.

$x_{1,2} = -\frac{b}{4a} - S \pm \frac12\sqrt{-4S^2 - 2p + \frac{q}{S}}$
$x_{3,4} = -\frac{b}{4a} + S \pm \frac12\sqrt{-4S^2 - 2p - \frac{q}{S}}$

where p and q are the coefficients of the second and of the first degree respectively in the associated depressed quartic

$p = \frac{8ac-3b^2}{8a^2}\qquad\qquad {\color{white}.}$
$q = \frac{b^3 - 4abc + 8a^2d}{8a^3}$

and where

$S = \frac{1}{2}\sqrt{-\frac23\ p+\frac{1}{3a}\left(Q + \frac{\Delta_0}{Q}\right)} \quad\qquad\ {\color{white}.}$ (if S = 0, see Special cases of the formula, below)
$Q\ =\ \sqrt[3]{\frac{\Delta_1 + \sqrt{\Delta_1^2 - 4\Delta_0^3}}{2}} \quad\qquad\qquad {\color{white}.}$ (if Q = 0, see Special cases of the formula, below)

with

$\Delta_0 = c^2 - 3bd + 12ae$
$\Delta_1 = 2c^3 - 9bcd + 27b^2 e + 27ad^2 - 72ace$

and

$\Delta_1^2-4\Delta_0^3 = - 27 \Delta\ ,$ where $\Delta$ is the aforementioned discriminant. The mathematical expressions of these last four terms are very similar to those of their cubic counterparts.

#### Special cases of the formula

If $\Delta > 0,$ using $Q$ may prove inconvenient, since its value is now a complex number. However, if all four roots are real, the value of $S$ is also real, and it would be simpler to express it with the aid of trigonometric functions, as follows:

$S = \frac{1}{2}\sqrt{-\frac23\ p+\frac{2}{3a}\sqrt{\Delta_0}\cos\frac{\phi}{3}}$

where

$\phi = \arccos\left(\frac{\Delta_1}{2\sqrt{\Delta_0^3}}\right).$

If $\Delta \neq 0$ and $\Delta_0 = 0,$ the sign of $\sqrt{\Delta_1^2 - 4 \Delta_0^3}=\sqrt{\Delta_1^2}$ has to be chosen to have $Q \neq 0,$ that is one should define $\sqrt{\Delta_1^2}$ as $\Delta_1,$ maintaining the sign of $\Delta_1.$

If $S = 0,$ then one must change the choice of the cubic root in $Q$ for having $S \neq 0.$ This is always possible except if the quartic may be factored into $\left(x+\tfrac{b}{4a}\right)^4.$ The result is then correct, but misleading hiding the fact that no cubic root is needed in this case. In fact this case may occur only if the numerator of $q$ is zero, and the associated depressed quartic is biquadratic; it may thus be solved by the method described below.

If $\Delta = 0$ and $\Delta_0 = 0,$ and thus also $\Delta_1 = 0,$ at least three roots are equal, and the roots are rational functions of the coefficients.

If $\Delta=0$ and $\Delta_0 \neq 0,$ the above expression for the roots is correct but misleading, hiding the fact that the polynomial is reducible and no cubic root is needed to represent the roots.

### Simpler cases

#### Reducible quartics

Consider the general quartic

$Q(x) = a_4x^4+a_3x^3+a_2x^2+a_1x+a_0.$

It is reducible if Q=RS, where R and S are non-constant polynomials with rational coefficients (or more generally with coefficients in the same field as the coefficients of Q). There are two ways to write such a factorization: Either

$Q(x) = (x-x_1)(b_3x^3+b_2x^2+b_1x+b_0)$

or

$Q(x) = (c_2x^2+c_1x+c_0)(d_2x^2+d_1x+d_0).$

In either case, the roots of Q are the roots of the factors, which may be computed by solving quadratic or cubic equations.

Detecting such factorizations can be done by using the factor function of every computer algebra system. But, in many cases, it may be done by hand-written computation. In the preceding section, we have already seen that the polynomial is always reducible if its discriminant $\Delta$ is zero (this is true for polynomials of every degree).

A very special case of the first case of factorization is when a0=0. This implies that x1=0 is a first root, b3=a4, b2=a3, b1=a2, b0=a1, and the other roots may be computed by solving a cubic equation.

If $a_4+a_3+a_2+a_1+a_0=0,$ then $Q(1)=0$ and we have a factorization of the first kind with x1=1. Similarly, if $a_4-a_3+a_2-a_1+a_0=0,$ then $Q(-1)=0$ and we have a factorization of the first kind with x1=-1.

Once a root x1 is known, the second factor of the factorization of the first kind is the quotient of the Euclidean division of Q by x-x1. It is

$a_4x^3+(a_4x_1+a_3)x^2+(a_4x_1^2+a_3x_1+a_2)x+a_4x_1^3+a_3x_1^2+a_2x_1+a_1.$

If $a_0, a_1, a_2, a_3, a_4$ are small integers a factorization of the first kind is easy to detect: if $x_1=\frac{p}{q}\ ,$ with p and q coprime integers, then q divides evenly a4, and p divides evenly a0. Thus, computing $Q\left(\frac{p}{q}\right)$ for every possible values of p and q allows to find the rational roots, if any.

In the case of two quadratic factors or of large integer coefficients, the factorization is harder to compute, and, in general, it is better to use the factor function of a computer algebra system (see polynomial factorization for a description of the algorithms that are involved).

If $a_3=a_1=0,\,$ then the biquadratic function

$Q(x) = a_4x^4+a_2x^2+a_0\,\!$

defines a biquadratic equation, which is easy to solve.

Let $z=x^2.\,$ Then Q becomes a quadratic q in $z,\,$

$q(z) = a_4z^2+a_2z+a_0.\,\!$

Let $z_+\,$ and $z_-\,$ be the roots of q. Then the roots of our quartic Q are

\begin{align} x_1&=+\sqrt{z_+}, \\ x_2&=-\sqrt{z_+}, \\ x_3&=+\sqrt{z_-}, \\ x_4&=-\sqrt{z_-}. \end{align}

#### Quasi-symmetric equations

$a_0x^4+a_1x^3+a_2x^2+a_1 m x+a_0 m^2=0 \,$

Steps:

1. Divide by x 2.
2. Use variable change z = x + m/x.

### Converting to a depressed quartic

For solving purpose, it is generally better to convert the quartic into a depressed quartic by the following simple change of variable. All formulas are simpler and some methods work only in this case. The roots of the original quartic are easily recovered from that of the depressed quartic by the reverse change of variable.

Let

$a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0 = 0 \qquad\qquad(1')$

be the general quartic equation we want to solve.

Dividing by a4, provides the equivalent equation

$x^4 + a x^3 + b x^2 + c x + d = 0,$

with

$a=\frac{a_3}{a_4},\quad b=\frac{a_2}{a_4},\quad c=\frac{a_1}{a_4},\quad d=\frac{a_0}{a_4}.$

Substituting x by $y-\frac{a_3}{4a_4}$ gives, after a simple term regrouping, the equation

$y^4+py^2+qy+r=0,$

where

\begin{align} p=&\frac{8b-3a^2}{8} &=&\frac{8a_2a_4-3a_3^2}{8a_4^2}\\ q=&\frac{a^3-4ab+8c}{8} &=&\frac{a_3^3-4a_2a_3a_4+8a_1a_4^2}{8a_4^3}\\ r=&\frac{-3a^4+256d-64ca+16a^2b}{256}&=&\frac{-3a_3^4+256a_0a_4^3-64a_1a_3a_4^2+16a_2a_3^2a_4}{256a_4^4} \end{align}

If y1, y2, y3, y4 are the roots of this depressed quartic, then the roots of the original quartic are $y_1-\frac{a}{4}=y_1-\frac{a_3}{4a_4},\; y_2-\frac{a}{4}=y_2-\frac{a_3}{4a_4},\;y_3-\frac{a}{4}=y_3-\frac{a_3}{4a_4},\;y_4-\frac{a}{4}=y_4-\frac{a_3}{4a_4}.$

### Ferrari's solution

As explained in the preceding section, we may start with a depressed quartic equation

$u^4 + \alpha u^2 + \beta u + \gamma = 0. \qquad \qquad (1)$

This depressed quartic can be solved by means of a method discovered by Lodovico Ferrari. The depressed equation may be rewritten (this is easily verified by expanding the square and regrouping all terms in the left-hand side)

$(u^2 + \alpha)^2 = \alpha u^2 + \alpha^2- \beta u - \gamma. \qquad \qquad (2)$

Then, we introduce a variable y into the factor on the left-hand side by adding $2u^2y+2\alpha y + y^2$ to both sides. After regrouping the coefficients of the power of u in the right-hand side, this gives the equation

$(u^2 + \alpha + y)^2 = (\alpha + 2 y) u^2 - \beta u + (y^2 + 2 y \alpha + \alpha^2 - \gamma), \qquad \qquad (3)\,$

which is equivalent to the original equation, whichever value is given to y.

As the value of y may be arbitrarily chosen, we will choose it in order to get a perfect square in the right-hand side. This implies that the discriminant in u of this quadratic equation is zero, that is y is a root of the equation

$(-\beta)^2 - 4 (2 y + \alpha) (y^2 + 2 y \alpha + \alpha^2 - \gamma) = 0.\,$

which may be rewritten

$y^3 + {5 \over 2} \alpha y^2 + (2 \alpha^2 - \gamma) y + \left( {\alpha^3 \over 2} - {\alpha \gamma \over 2} - {\beta^2 \over 8} \right) = 0. \qquad \qquad (4)$

The value of y may thus be obtained from the formulas provided in the article Cubic equation.

When y is a root of equation (4), the right-hand side of equation (3) the square of

$\left(u\sqrt{\alpha + 2 y}-\frac{\beta}{2\sqrt{\alpha + 2 y}}\right)^2.$

However, this induces a division by zero if $\alpha + 2 y=0.$ This implies $\beta=0,$ and thus that the depressed equation is bi-quadratic, and may be solved by an easier method (see above). This was not a problem at the time of Ferrari, when one solved only explicitly given equations with numeric coefficients. For a general formula that is always true, one thus need to choose a root of the cubic equation such that $\alpha + 2 y \not=0.$ This is always possible unless for the depressed equation x4=0.

Now, if y is a root of the cubic equation such that $\alpha + 2 y \not=0,$ equation (3) may be rewritten

$\left(u^2 + \alpha + y+u\sqrt{\alpha + 2 y}-\frac{\beta}{2\sqrt{\alpha + 2 y}}\right) \left(u^2 + \alpha + y - u\sqrt{\alpha + 2 y}+\frac{\beta}{2\sqrt{\alpha + 2 y}}\right)=0,$

and the equation is easily solved by applying to each factor the formula for quadratic equations. Solving them we may write the four roots as

$u={\pm_1\sqrt{\alpha + 2 y} \pm_2 \sqrt{-\left(3\alpha + 2y \pm_1 {2\beta \over \sqrt{\alpha + 2 y}} \right)} \over 2},$

where $\pm_1$ and $\pm_2$ denote either + or -. As the two occurrences of $\pm_1$ must denote the same sign, this leave four possibilities, one for each root.

Therefore the solutions of the original quartic equation are

$x=-{a_3 \over 4a_4} + {\pm_1\sqrt{\alpha + 2 y} \pm_2 \sqrt{-\left(3\alpha + 2y \pm_1 {2\beta \over \sqrt{\alpha + 2 y}} \right)} \over 2}. \qquad\qquad (8')$

### Solving by factoring into quadratics

One can solve a quartic by factoring it into a product of two quadratics.[10] Let

$\begin{array}{lcl} 0 = x^4 + ax^3 + bx^2 + cx + d & = & (x^2 + px + q)(x^2 + rx + s) \\ & = & x^4 + (p + r)x^3 + (q + s + pr)x^2 + (ps + qr)x + qs \end{array}$

By equating coefficients, this results in the following set of simultaneous equations:

$\begin{array}{lcl} a & = & p + r \\ b & = & q + s + pr \\ c & = & ps + qr \\ d & = & qs \end{array}$

This can be simplified by starting again with a depressed quartic where $a = 0$, which can be obtained by substituting $(x - a/4)$ for $x$, then $r = -p$, and:

$\begin{array}{lcl} b + p^2 & = & s + q \\ c & = & (s - q) p \\ d & = & sq \end{array}$

It's now easy to eliminate both $s$ and $q$ by doing the following:

$\begin{array}{lcl} p^2(b + p^2)^2 - c^2 & = & p^2(s + q)^2 - p^2(s - q)^2 \\ & = & 4p^2sq \\ & = & 4p^2d \end{array}$

If we set $P = p^2$, then this equation turns into the resolvent cubic equation

$P^3 + 2bP^2 + (b^2 - 4d)P - c^2 = 0\,$

which is solved elsewhere. Then, if p is a square root of a non-zero root of this resolvent (such a non zero root exists except for the quartic x4, which is trivially factored),

$\begin{array}{lcl} r & = & -p \\ 2s & = & b + p^2 + c/p \\ 2q & = & b + p^2 - c/p \end{array}$

The symmetries in this solution are easy to see. There are three roots of the cubic, corresponding to the three ways that a quartic can be factored into two quadratics, and choosing positive or negative values of $p$ for the square root of $P$ merely exchanges the two quadratics with one another.

The above solution shows that the quartic polynomial with a zero coefficient on the cubic term is factorable into quadratics with rational coefficients if and only if either the resolvent cubic $P^3 + 2bP^2 + (b^2 - 4d)P - c^2\,$ has a non-zero root which is the square of a rational, or $b^2-4d$ is the square of rational and c=0; this can readily be checked using the rational root test.

### Solving by Lagrange resolvent

The symmetric group S4 on four elements has the Klein four-group as a normal subgroup. This suggests using a resolvent cubic whose roots may be variously described as a discrete Fourier transform or a Hadamard matrix transform of the roots; see Lagrange resolvents for the general method. Denote by xi, for i from 0 to 3, the four roots roots of

$x^4 + ax^3 + bx^2 + cx + d = 0\qquad (1)$

If we set

\begin{align} s_0 &= \tfrac12(x_0 + x_1 + x_2 + x_3), \\ s_1 &= \tfrac12(x_0 - x_1 + x_2 - x_3), \\ s_2 &= \tfrac12(x_0 + x_1 - x_2 - x_3), \\ s_3 &= \tfrac12(x_0 - x_1 - x_2 + x_3), \end{align}

then since the transformation is an involution we may express the roots in terms of the four si in exactly the same way. Since we know the value s0 = -a/2, we only need the values for s1, s2 and s3. These are the roots of the polynomial

$(s^2 - s_1^2)(s^2-s_2^2)(s^2-s_3^2).\qquad (2)$

Substituting the si by their values in term of the xi, this polynomial may be expanded in a polynomial in s whose coefficients are symmetric polynomials in the xi. By the fundamental theorem of symmetric polynomials, these coefficients may be expressed as polynomials in the coefficients of the monic quartic. If, for simplification, we suppose that the quartic is depressed, that is a=0, this results in the polynomial

${s}^{6}+2b\,{s}^{4}+({b}^{2}-4c)\,{s}^{2}-{c}^{2}\qquad(3)$

This polynomial is of degree six, but only of degree three in s2, and so the corresponding equation is solvable by the method described in the article Cubic function. By substituting the roots in the expression of the xi in terms of the si, we obtain expression for the roots. In fact we obtain, apparently, several expressions, depending on the numbering of the roots of the cubic polynomial and of the signs given to their square roots. All these different expressions may be deduced from one of them by simply changing the numbering of the xi.

These expressions are unnecessarily complicated, involving the cubic roots of unity, which can be avoided as follows. If s is any non-zero root of (3), and if we set

$F_1 = {x}^{2}+sx+ \frac{b}{2} + \frac{s^2}{2} - \frac{c}{2p}$
$F_2 ={x}^{2}-sx+ \frac{b}{2} + \frac{s^2}{2} + \frac{c}{2p}$

then

$F_1 F_2 = x^4 + bx^2 + cx + d$

We therefore can solve the quartic by solving for s and then solving for the roots of the two factors using the quadratic formula.

Note that this gives exactly the same formula for the roots as the preceding section.

### Solving with algebraic geometry

An alternative solution using algebraic geometry is given in (Faucette 1996), and proceeds as follows (more detailed discussion in reference). In brief, one interprets the roots as the intersection of two quadratic curves, then finds the three reducible quadratic curves (pairs of lines) that pass through these points (this corresponds to the resolvent cubic, the pairs of lines being the Lagrange resolvents), and then use these linear equations to solve the quadratic.

The four roots of the depressed quartic $x^4+px^2+qx+r=0$ may also be expressed as the x coordinates of the intersections of the two quadratic equations $y^2+py+qx+r=0,$ $y-x^2=0;$ i.e., using the substitution $y=x^2;$ that two quadratics intersect in four points is an instance of Bézout's theorem. Explicitly, the four points are $P_i := (x_i,x_i^2)$ for the four roots $x_i$ of the quartic.

These four points are not collinear because they lie on the irreducible quadratic $y=x^2,$ and thus there is a 1-parameter family of quadratics (a pencil of curves) passing through these points. Writing the projectivization of the two quadratics as quadratic forms in three variables:

\begin{align} F_1(X,Y,Z) &:= Y^2 + pYZ + qXZ + rZ^2,\\ F_2(X,Y,Z) &:= YZ - X^2 \end{align}

the pencil is given by the forms $\lambda F_1 + \mu F_2$ for any point $[\lambda,\mu]$ in the projective line – in other words, where $\lambda$ and $\mu$ are not both zero, and multiplying a quadratic form by a constant does not change its quadratic curve of zeros.

This pencil contains three reducible quadratics, each corresponding to a pair of lines, each passing through two of the four points, which can be done $\textstyle{\binom{4}{2} = 6}$ different ways. Denote these $Q_1 = L_{12} + L_{34},$ $Q_2 = L_{13} + L_{24},$ $Q_3 = L_{14} + L_{23}.$ Given any two of these, their intersection is exactly the four points.

The reducible quadratics, in turn, may be determined by expressing the quadratic form $\lambda F_1 + \mu F_2$ as a 3×3 matrix: reducible quadratics correspond to this matrix being singular, which is equivalent to its determinant being zero, and the determinant is a homogeneous degree three polynomial in $\lambda$ and $\mu,$ and corresponds to the resolvent cubic.

## References

1. ^
2. ^ Depman (1954), Rasskazy o matematike (in Russian), Leningrad: Gosdetizdat
3. ^ P. Beckmann (1971). A history of π. Macmillan. p. 80.
4. ^ P. Beckmann (1971). A history of π. Macmillan. p. 191.
5. ^ P. Zoll (1989). "Letter to the Editor". American Mathematical Monthly 96 (8): 709–710. JSTOR 2324719.
6. ^ Stewart, Ian, Galois Theory, Third Edition (Chapman & Hall/CRC Mathematics, 2004)
7. ^ Rees, E. L. (1922). "Graphical Discussion of the Roots of a Quartic Equation". The American Mathematical Monthly 29 (2): 51–55.
8. ^ Lazard, D. (1988). "Quantifier elimination: Optimal solution for two classical examples". Journal of Symbolic Computation 5: 261–266. doi:10.1016/S0747-7171(88)80015-4.
9. ^ http://planetmath.org/QuarticFormula, PlanetMath, quartic formula, 21st October 2012
10. ^ Brookfield, G. (2007). "Factoring quartic polynomials: A lost art". Mathematics Magazine 80 (1): 67–70.