# Quartic function

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 sides; 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).

It is reported that even earlier, in 1486, Spanish mathematician Paolo Valmes was burned at the stake for claiming to have solved the quartic equation. Inquisitor General Tomás de Torquemada allegedly told him that it was the will of God that such a solution be inaccessible to human understanding.[2] However, attempts to find corroborating evidence for this story, or for the existence of Paolo Valmes, have not succeeded.[3]

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.[4]

## 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 and during ray-tracing against surfaces such as quadric or tori surfaces, which are the next level beyond the sphere and developable surfaces.

Another frequent generator of quartics is the intersection of two ellipses.

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

The discriminant of the quartic $x^4+ax^3+bx^2+cx+d$ is

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

There are three possible cases:

• If $\Delta > 0$ the equation's four roots are either all real or all complex.
• If $\Delta = 0$ then either the polynomial has a multiple rational root, or it is the square of a quadratic polynomial.
• If $\Delta < 0$ the equation has two real roots and two complex roots.

### 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.[5]

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

$a_4x^4+a_3x^3+a_2x^2+a_1x+a_0=0 \,$

with $a_4 \neq 0$ are equal to those of the corresponding monic polynomial

$x^4+ax^3+bx^2+cx+d=0 \,$

where

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

The roots in terms of these four coefficients are given by:

$\begin{cases} x_{1,2} = -\tfrac14\ (a\ +\ 2\sqrt{u\ +\ v}\ \pm\ 2\sqrt{2u\ -\ v\ +\ w}\ ) \\ \\ x_{3,4} = -\tfrac14\ (a\ -\ 2\sqrt{u\ +\ v}\ \pm\ 2\sqrt{2u\ -\ v\ -\ w}\ ) \end{cases}$

where

$u = \frac{3a^2 - 8b}{12}$
$v = \frac{Q^2 + \Delta_0}{3 Q}$
$w = \frac{a^3 - 4ab + 8c}{4\sqrt{u\ +\ v}}$

and where

$Q = \sqrt[3]{\frac{\Delta_1 + \sqrt{\Delta_1^2 - 4\Delta_0^3}}{2}} \qquad \qquad {\color{white}.}$ (see below)

with

$\Delta_0 = b^2 - 3ac + 12d$
$\Delta_1 = 2b^3 - 9abc + 27a^2 d + 27c^2 - 72bd$

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 almost identical with those of their cubic counterparts.

### Special cases

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 $\Delta = 0$ and $\ \Delta_0 = 0$, 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.

#### Reducible quartics

Consider the general quartic

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

It is irreducible 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 quadric 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 $Q(\frac{p}{q})$ 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 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 recoverd 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, and substituting x by $y-\frac{a_3}{4a_4}$ gives, after a simple term regrouping, the equation

$y^4+px^2+qx+r=0,$

where

\begin{align} p&=\frac{8a_2a_4-3a_3^2}{8a_4^2}\\ q&=\frac{a_3^3-4a_2a_3a_4+8a_1a_4^2}{8a_4^3}\\ r&=\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 root of this depressed quartic, then the roots of the original quartic are $y_1-\frac{a_3}{4a_4},\; y_2-\frac{a_3}{4a_4},\;y_3-\frac{a_3}{4a_4},\;y_4-\frac{a_3}{4a_4}.$

### The general case, along Ferrari's lines

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

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

#### Ferrari's solution

The depressed quartic can be solved by means of a method discovered by Lodovico Ferrari. Once the depressed quartic has been obtained, the next step is to add the valid identity

$(u^2 + \alpha)^2 - u^4 - 2 \alpha u^2 = \alpha^2\,$

to equation (1), yielding

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

The effect has been to fold up the u4 term into a perfect square: (u2 + α)2. The second term, αu2 did not disappear, but its sign has changed and it has been moved to the right side.

The next step is to insert a variable y into the perfect square on the left side of equation (2), and a corresponding 2y into the coefficient of u2 in the right side. To accomplish these insertions, the following valid formulas will be added to equation (2),

$\begin{matrix} (u^2+\alpha+y)^2-(u^2+\alpha)^2 & = & 2y(u^2+\alpha)+ y^2\\ \\ & = & 2yu^2+2y\alpha+y^2, \end{matrix}$

and

$0 = (\alpha + 2 y) u^2 - 2 y u^2 - \alpha u^2\,$

These two formulas, added together, produce

$(u^2 + \alpha + y)^2 - (u^2 + \alpha)^2 = (\alpha + 2 y) u^2 - \alpha u^2 + 2 y \alpha + y^2 \qquad \qquad (y-\hbox{insertion})\,$

which added to equation (2) produces

$(u^2 + \alpha + y)^2 + \beta u + \gamma = (\alpha + 2 y) u^2 + (2 y \alpha + y^2 + \alpha^2).\,$

This is equivalent to

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

The objective now is to choose a value for y such that the right side of equation (3) becomes a perfect square. This can be done by letting the discriminant of the quadratic function become zero. To explain this, first expand a perfect square so that it equals a quadratic function:

$(s u + t)^2 = (s^2) u^2 + (2 s t) u + (t^2).\,$

The quadratic function on the right side has three coefficients. It can be verified that squaring the second coefficient and then subtracting four times the product of the first and third coefficients yields zero:

$(2 s t)^2 - 4 (s^2) (t^2) = 0.\,$

Therefore to make the right side of equation (3) into a perfect square, the following equation must be solved:

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

Multiply the binomial with the polynomial,

$\beta^2 - 4 (2 y^3 + 5 \alpha y^2 + (4 \alpha^2 - 2 \gamma) y + (\alpha^3 - \alpha \gamma)) = 0\,$

Divide both sides by −4, and move the −β2/4 to the right,

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

This is a cubic equation for y. Divide both sides by 2,

$y^3 + {5 \over 2} \alpha y^2 + (2 \alpha^2 - \gamma) y + \left( {\alpha^3 \over 2} - {\alpha \gamma \over 2} - {\beta^2 \over 8} \right) = 0. \qquad \qquad (4)$
##### Conversion of the nested cubic into a depressed cubic

Equation (4) is a cubic equation nested within the quartic equation. It must be solved to solve the quartic. To solve the cubic, first transform it into a depressed cubic by means of the substitution

$y = v - {5 \over 6} \alpha.$

Equation (4) becomes

$\left( v - {5 \over 6} \alpha \right)^3 + {5 \over 2} \alpha \left( v - {5 \over 6} \alpha \right)^2 + (2 \alpha^2 - \gamma) \left( v - {5 \over 6} \alpha \right) + \left( {\alpha^3 \over 2} - {\alpha \gamma \over 2} - {\beta^2 \over 8} \right) = 0.$

Expand the powers of the binomials,

$\left( v^3 - {5 \over 2} \alpha v^2 + {25 \over 12} \alpha^2 v - {125 \over 216} \alpha^3 \right) + {5 \over 2} \alpha \left( v^2 - {5 \over 3} \alpha v + {25 \over 36} \alpha^2 \right) + (2 \alpha^2 - \gamma) \left( v - {5 \over 6} \alpha \right) + \left( {\alpha^3 \over 2} - {\alpha \gamma \over 2} - {\beta^2 \over 8} \right) = 0.$

Distribute, collect like powers of v, and cancel out the pair of v2 terms,

$v^3 + \left( - {\alpha^2 \over 12} - \gamma \right) v + \left( - {\alpha^3 \over 108} + {\alpha \gamma \over 3} - {\beta^2 \over 8} \right) = 0.$

This is a depressed cubic equation.

Relabel its coefficients,

$P = - {\alpha^2 \over 12} - \gamma,$
$Q = - {\alpha^3 \over 108} + {\alpha \gamma \over 3} - {\beta^2 \over 8}.$

The depressed cubic now is

$v^3 + P v + Q = 0. \qquad \qquad (5)$
##### Solving the nested depressed cubic

The solutions (any solution will do, so pick any of the three complex roots) of equation (5) are computed as (see Cubic equation)

$y = - {5 \over 6} \alpha + U + V \qquad \qquad (6)$

where

$U=\sqrt[3]{-{Q\over 2}\pm \sqrt{{Q^{2}\over 4}+{P^{3}\over 27}}}$

and V is computed according to the two defining equations $-U \mbox{ }^3-V \mbox{ }^3=Q$ and $-3UV=P$, so

$V=\begin{cases} -\frac{P}{3U}&\text{ if }U\ne 0\text{ and }\\ -\sqrt[3]{Q}&\text{ if }U=0\ . \end{cases}$
##### Folding the second perfect square

With the value for y given by equation (6), it is now known that the right side of equation (3) is a perfect square of the form

$(s^2)u^2+(2st)u+(t^2) = (s^2)\left(u - {-(2st) \over 2(s^2)}\right)^2=\left[\sqrt{(s^2)}\left(u - {-(2st) \over 2(s^2)}\right)\right]^2,$
$(s^2)u^2+(2st)u+(t^2) = \left(\left(\sqrt{(s^2)}\right)u + {(2st) \over 2\sqrt{(s^2)}}\right)^2$
(This is correct for both signs of square root, as long as the same sign is taken for both square roots. A ± is redundant, as it would be absorbed by another ± a few equations further down this page.)

so that it can be folded:

$(\alpha + 2 y) u^2 + (- \beta) u + (y^2 + 2 y \alpha + \alpha^2 - \gamma ) = \left( \left(\sqrt{(\alpha + 2y)}\right)u + {(-\beta) \over 2\sqrt{(\alpha + 2 y)}} \right)^2$.
Note: If β ≠ 0 then α + 2y ≠ 0. If β = 0 then this would be a biquadratic equation, which we solved earlier.

Therefore equation (3) becomes

$(u^2 + \alpha + y)^2 = \left( \left(\sqrt{\alpha + 2 y}\right)u - {\beta \over 2\sqrt{\alpha + 2 y}} \right)^2 \qquad\qquad (7)$.

Equation (7) has a pair of folded perfect squares, one on each side of the equation. The two perfect squares balance each other.

If two squares are equal, then the sides of the two squares are also equal, as shown by:

$(u^2 + \alpha + y) = \pm\left( \left(\sqrt{\alpha + 2 y}\right)u - {\beta \over 2\sqrt{\alpha + 2 y}} \right) \qquad\qquad (7')$.

Collecting like powers of u produces

$u^2 + \left(\mp_s \sqrt{\alpha + 2 y}\right)u + \left( \alpha + y \pm_s {\beta \over 2\sqrt{\alpha + 2 y}} \right) = 0 \qquad\qquad (8)$.
Note: The subscript s of $\pm_s$ and $\mp_s$ is to note that they are dependent.

Equation (8) is a quadratic equation for u. Its solution is

$u={\pm_s\sqrt{\alpha + 2 y} \pm_t \sqrt{(\alpha + 2y) - 4(\alpha + y \pm_s {\beta \over 2\sqrt{\alpha + 2 y}})} \over 2}.$

Simplifying, one gets

$u={\pm_s\sqrt{\alpha + 2 y} \pm_t \sqrt{-\left(3\alpha + 2y \pm_s {2\beta \over \sqrt{\alpha + 2 y}} \right)} \over 2}.$

This is the solution of the depressed quartic, therefore the solutions of the original quartic equation are

$x=-{B \over 4A} + {\pm_s\sqrt{\alpha + 2 y} \pm_t \sqrt{-\left(3\alpha + 2y \pm_s {2\beta \over \sqrt{\alpha + 2 y}} \right)} \over 2}. \qquad\qquad (8')$
Remember: The two $\pm_s$ come from the same place in equation (7'), and should both have the same sign, while the sign of $\pm_t$ is independent.
##### Summary of Ferrari's method

Given the quartic equation

$A x^4 + B x^3 + C x^2 + D x + E = 0, \,$

its solution can be found by means of the following calculations:

$\alpha = - {3 B^2 \over 8 A^2} + {C \over A},$
$\beta = {B^3 \over 8 A^3} - {B C \over 2 A^2} + {D \over A},$
$\gamma = - {3 B^4 \over 256 A^4} + {C B^2 \over 16 A^3} - {B D \over 4 A^2} + {E \over A}.$

If $\,\beta=0,$ then

$x=-{B\over 4A}\pm_s\sqrt{-\alpha\pm_t\sqrt{\alpha^2-4\gamma}\over 2}\qquad \mbox{(for } \beta=0 \mbox{ only)}.$

Otherwise, continue with

$P = - {\alpha^2 \over 12} - \gamma,$
$Q = - {\alpha^3 \over 108} + {\alpha \gamma \over 3} - {\beta^2 \over 8},$
$R = -{Q\over 2} \pm \sqrt{{Q^{2}\over 4}+{P^{3}\over 27}},$

(either sign of the square root will do)

$U = \sqrt[3]{R},$

(there are 3 complex roots, any one of them will do)

$y = \begin{cases} - {5 \over 6} \alpha + U - \frac{P}{3U} & \text{if }U\ne 0\\ -{5\over 6} \alpha - \sqrt[3]{Q} & \text{if }U=0 \end{cases}$
$W=\sqrt{ \alpha + 2 y}$
$x = - {B \over 4 A} + { \pm_s W \mp_t \sqrt{-\left(3\alpha + 2 y \pm_s {2\beta\over W} \right) }\over 2 }.$

As stated above, Cardano credited Ferrari as the first to discover one of these labyrinthine solutions. The equation he solved was:

$\ x^4 + 6 x^2 - 60 x + 36 = 0$

which was already in depressed form. It has a pair of solutions that can be found with the set of formulas shown above.

##### Ferrari's solution in the special case of real coefficients

If the coefficients of the quartic equation are real then the nested depressed cubic equation (5) also has real coefficients, thus it has at least one real root.

Furthermore the cubic function $\ C(v) = v^3 + P v + Q,$ where P and Q are given by (5) has the properties that

$C\left({\alpha \over 3}\right) = {-\beta^2 \over 8} < 0 \$ and

$\lim_{v\to \infty}{C(v)}=\infty,$ where α and β are given by (1).

This means that (5) has a real root greater than $\alpha \over 3$, and therefore that (4) has a real root greater than $-\alpha \over 2$.

Using this root the term $\sqrt{\alpha + 2 y}$ in (8) is always real, which ensures that the two quadratic equations (8) have real coefficients.[6]

#### Obtaining alternative solutions by factoring out complex conjugate solutions

It could happen that one only obtained one solution through the seven formulae above, because not all four sign patterns are tried for four solutions, and the solution obtained is complex. It may also be the case that one is only looking for a real solution. Let x1 denote the complex solution. If all the original coefficients A, B, C, D and E are real — which should be the case when one desires only real solutions — then there is another complex solution x2, which is the complex conjugate of x1. If the other two roots are denoted as x3 and x4 then the quartic equation can be expressed as

$(x - x_1) (x - x_2) (x - x_3) (x - x_4) = 0, \,$

but this quartic equation is equivalent to the product of two quadratic equations:

$(x - x_1) (x - x_2) = 0 \qquad \qquad (9)$

and

$(x - x_3) (x - x_4) = 0. \qquad \qquad (10)$

Since

$x_2 = x_1^\star$

then

$\begin{matrix} (x-x_1)(x-x_2)&=&x^2-(x_1+x_1^\star)x+x_1x_1^\star\qquad\qquad\qquad\quad \\ &=&x^2-2\,\mathrm{Re}(x_1)x+[\mathrm{Re}(x_1)]^2+[\mathrm{Im}(x_1)]^2. \end{matrix}$

Let

$a = - 2 \, \mathrm{Re}(x_1),$
$b = \left[ \mathrm{Re}( x_1) \right]^{2} + \left[ \mathrm{Im}(x_1) \right]^{2}$

so that equation (9) becomes

$x^2 + a x + b = 0. \qquad \qquad (11)$

Also let there be (unknown) variables w and v such that equation (10) becomes

$x^2 + w x + v = 0. \qquad \qquad (12)$

Multiplying equations (11) and (12) produces

$x^4 + (a + w) x^3 + (b + w a + v) x^2 + (w b + v a) x + v b = 0. \qquad \qquad (13)$

Comparing equation (13) to the original quartic equation, it can be seen that

$a + w = {B \over A},$
$b + w a + v = {C \over A},$
$w b + v a = {D \over A},$

and

$v b = {E \over A}.$

Therefore

$w = {B \over A} - a = {B \over A} + 2 \mathrm{Re}(x_1),$
$v = {E \over A b} = {E \over A \left( [ \mathrm{Re}(x_1) ]^2 + [ \mathrm{Im}(x_1) ]^2 \right) }.$

Equation (12) can be solved for x yielding

$x_3 = {-w + \sqrt{w^2 - 4 v} \over 2},$
$x_4 = {-w - \sqrt{w^2 - 4 v} \over 2}.$

These two solutions are the desired real solutions if real solutions exist.

### Alternative methods

One can solve a quartic by factoring it into a product of two quadratics.[7] Let

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

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

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

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

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

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

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

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

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

which is solved elsewhere. Then:

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

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 the resolvent cubic $P^3 + 2cP^2 + (c^2 - 4e)P - d^2\,$ has a root which is the square of a rational; this can readily be checked using the rational root test.

#### Galois theory and factorization

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. Suppose ri for i from 0 to 3 are roots of

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

If we now set

\begin{align} s_0 &= \tfrac12(r_0 + r_1 + r_2 + r_3), \\ s_1 &= \tfrac12(r_0 - r_1 + r_2 - r_3), \\ s_2 &= \tfrac12(r_0 + r_1 - r_2 - r_3), \\ s_3 &= \tfrac12(r_0 - r_1 - r_2 + r_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 = -b/2, we really only need the values for s1, s2 and s3. These we may find by expanding the polynomial

$(z^2 - s_1^2)(z^2-s_2^2)(z^2-s_3^2)\qquad (2)$

which if we make the simplifying assumption that b=0, is equal to

${z}^{6}+2c\,{z}^{4}+({c}^{2}-4e)\,{z}^{2}-{d}^{2}\qquad(3)$

This polynomial is of degree six, but only of degree three in z2, and so the corresponding equation is solvable. By trial we can determine which three roots are the correct ones, and hence find the solutions of the quartic.

We can remove any requirement for trial by using a root of the same resolvent polynomial for factoring; if w is any root of (3), and if

$F_1 = {x}^{2}+wx+1/2\,{w}^{2}+1/2\,c-1/2\,{\frac {{c}^{2}w}{d}}-1/2\,{\frac {{w}^{5}}{d}}-{\frac {c{w}^{3}}{d}}+2\,{\frac {ew}{d}}$
$F_2 = {x}^{2}-wx+1/2\,{w}^{2}+1/2\,c+1/2\,{\frac {{w}^{5}}{d}}+{\frac {c{w}^{3}}{d}}-2\,{\frac {ew}{d}}+1/2\,{\frac {{c}^{2}w}{d}}$

then

$F_1 F_2 = x^4 + cx^2 + dx + e\,\,(4)$

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

#### 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.