Quartic equation

In mathematics, a quartic equation is the result of setting a quartic function equal to zero. An example of a quartic equation is the equation

0 qquad qquad

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

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: y^3 + {5 over 2} lpha y^2 + (2 lpha^2 - gamma) y + left( {lpha^3 over 2} - {lpha gamma over 2} - {eta^2 over 8} ight) = 0. qquad qquad (4)

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Conversion of the nested cubic into a depressed cubic

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

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: y = v - {5 over 6} lpha.

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Equation (4) becomes

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: left( v - {5 over 6} lpha ight)^3 + {5 over 2} lpha left( v - {5 over 6} lpha ight)^2 + (2 lpha^2 - gamma) left( v - {5 over 6} lpha ight) + left( {lpha^3 over 2} - {lpha gamma over 2} - {eta^2 over 8} ight) = 0.

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Expand the powers of the binomials,

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: left( v^3 - {5 over 2} lpha v^2 + {25 over 12} lpha^2 v - {125 over 216} lpha^3 ight) + {5 over 2} lpha left( v^2 - {5 over 3} lpha v + {25 over 36} lpha^2 ight) + (2 lpha^2 - gamma) v - {5 over 6} lpha (2 lpha^2 - gamma ) + left( {lpha^3 over 2} - {lpha gamma over 2} - {eta^2 over 8} ight) = 0.

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

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: v^3 + left( - {lpha^2 over 12} - gamma ight) v + left( - {lpha^3 over 108} + {lpha gamma over 3} - {eta^2 over 8} ight) = 0.

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This is a depressed cubic equation.

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Relabel its coefficients,

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: P = - {lpha^2 over 12} - gamma,

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: Q = - {lpha^3 over 108} + {lpha gamma over 3} - {eta^2 over 8}.

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The depressed cubic now is

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: v^3 + P v + Q = 0. qquad qquad (5)

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Solving the nested depressed cubic

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

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:::let U=sqrt{{Qover 2}pm sqrt{{Q^{2}over 4}+{P^{3}over 27}}}

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::::(taken from Cubic equation)

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:v = {Pover 3U} - U

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therefore the solution of the original nested cubic is

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:y = - {5 over 6} lpha + {Pover 3U} - U qquad qquad (6)

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::Remember 1: P=0 Longleftarrow {Qover 2} + sqrt{{Q^{2}over 4}+{P^{3}over 27}}=0

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::Remember 2: lim_{P o 0}{P over sqrt{{Qover 2} + sqrt{{Q^{2}over 4}+{P^{3}over 27}}}}=0

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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

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:(s^2)u^2+(2st)u+(t^2) = left(left(sqrt{(s^2)} ight)u + {(2st) over 2sqrt{(s^2)}} ight)^2

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::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.

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so that it can be folded:

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: (lpha + 2 y) u^2 + (- eta) u + (y^2 + 2 y lpha + lpha^2 - gamma ) = left( left(sqrt{(lpha + 2y)} ight)u + {(-eta) over 2sqrt{(lpha + 2 y)}} ight)^2.

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::Note: If β ≠ 0 then α + 2y ≠ 0. If β = 0 then this would be a biquadratic equation, which we solved earlier.

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Therefore equation (3) becomes

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:(u^2 + lpha + y)^2 = left( left(sqrt{lpha + 2 y} ight)u - {eta over 2sqrt{lpha + 2 y}} ight)^2 qquadqquad (7).

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

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

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:(u^2 + lpha + y) = pmleft( left(sqrt{lpha + 2 y} ight)u - {eta over 2sqrt{lpha + 2 y}} ight) qquadqquad (7').

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Collecting like powers of u produces

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:u^2 + left(mp_s sqrt{lpha + 2 y} ight)u + left( lpha + y pm_s {eta over 2sqrt{lpha + 2 y}} ight) = 0 qquadqquad (8).

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::Note: The subscript s of pm_s and mp_s is to note that they are dependent.

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

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:u={pm_ssqrt{lpha + 2 y} pm_t sqrt{(lpha + 2y) - 4(lpha + y pm_s {eta over 2sqrt{lpha + 2 y}})} over 2}.

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Simplifying, one gets

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:u={pm_ssqrt{lpha + 2 y} pm_t sqrt{-left(3lpha + 2y pm_s {2eta over sqrt{lpha + 2 y}} ight)} over 2}.

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

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:x=-{B over 4A} + {pm_ssqrt{lpha + 2 y} pm_t sqrt{-left(3lpha + 2y pm_s {2eta over sqrt{lpha + 2 y}} ight)} over 2}. qquadqquad (8')

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::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.

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Summary of Ferrari's method

Given the quartic equation

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

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

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: lpha = - {3 B^2 over 8 A^2} + {C over A},

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: eta = {B^3 over 8 A^3} - {B C over 2 A^2} + {D over A},

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: gamma = {-3 B^4 over 256 A^4} + {C B^2 over 16 A^3} - {B D over 4 A^2} + {E over A},

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::if eta=0 solve u^4+lpha u^2 + gamma = 0 and substitute x=u-{Bover 4A} finding the roots

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:::x=-{Bover 4A}pm_ssqrt{-lphapm_tsqrt{lpha^2-4gamma}over 2},qquadeta=0.

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: P = - {lpha^2 over 12} - gamma,

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: Q = - {lpha^3 over 108} + {lpha gamma over 3} - {eta^2 over 8},

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: R = {Qover 2} pm sqrt{{Q^{2}over 4}+{P^{3}over 27}}, (either sign of the square root will do, as long as R does not disappear unnecessarily; in the case of P=0 we want R=Q)

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: U = sqrt{R}, (there are 3 complex roots, any one of them will do)

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: y = - {5 over 6} lpha -U + egin{cases}U=0 & o 0\U e 0 & o {Pover 3U}end{cases},

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: x = - {B over 4 A} + { pm_s sqrt{ lpha + 2 y} pm_t sqrt{-left(3lpha + 2 y pm_s {2etaoversqrt{lpha +2y}} ight) }over 2 }.

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::The two ±s must have the same sign, the ±t is independent. To get all roots, find x for ±s,±t = +,+ and for +,− and for −,+ and for −,−. Double roots will be given twice, triple roots 3 times and quadruple roots would be given 4 times (although then β = 0, which is a special case). The order of the roots depends on which cubic root U one chose. (see note for (8) vis-à-vis (8'))

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Quod Erat Faciendum.

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There are other methods of solving the quartic equations, perhaps more optimal. Ferrari was the first to discover one of these labyrinthine solutions. The equation which he solved was

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: x^4 + 6 x^2 - 60 x + 36 = 0

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

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Obtaining alternative solutions the hard way

It could happen that one only obtained one solution through the seven formulae above, because one doesn't like trying all four sign patterns to get all four solutions, and the solution one 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

Related Topics:
Complex - Complex conjugate

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: (x - x_1) (x - x_2) (x - x_3) (x - x_4) = 0,

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

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: (x - x_1) (x - x_2) = 0 qquad qquad (9)

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and

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: (x - x_3) (x - x_4) = 0. qquad qquad (10)

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Since

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: x_2 = x_1^star

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then

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:

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egin{matrix}

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(x-x_1)(x-x_2)&=&x^2-(x_1+x_1^star)x+x_1x_1^starqquadqquadqquadquad

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\

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&=&x^2-2,mathrm{Re}(x_1)x+^2+^2.

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end{matrix}

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Let

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: a = - 2 , mathrm{Re}(x_1),

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: b = ^2 + ^2

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so that equation (9) becomes

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

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

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

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Multiplying equations (11) and (12) produces

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

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

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: a + w = {B over A},

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: b + w a + v = {C over A},

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: w b + v a = {D over A},

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and

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: v b = {E over A}.

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Therefore

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: w = {B over A} - a = {B over A} + 2 mathrm{Re}(x_1),

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: v = {E over A b} = {E over A left(

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^2 + ^2 ight) }.

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Equation (12) can be solved for x yielding

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: x_3 = {-w + sqrt{w^2 - 4 v} over 2},

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: x_4 = {-w - sqrt{w^2 - 4 v} over 2}.

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One of these two solutions should be the desired real solution.

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Alternative methods

Reduction to a biquadratic

We may find the roots of

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:x^4 + cx^2 + dx + e = 0 qquadqquad (1) (see Quartic equation#Converting to a depressed quartic)

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by converting it to a biquadratic equation by means of a Tschirnhaus transformation.

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If y = x^2 + px + q, we may set p to be a root of

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:dp^3 + (4e-c^2)p^2 - 2cdp-d^2 = 0 (see Cubic equation)

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and

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:q to be cover 2.

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This transforms the equation to

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:y^4+(-c^2/2+3pd+2e+p^2c)y^2+

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:(48p^2ced^2-128pce^2d-4c^3p^2d^2+32pc^3ed-128p^2c^2e^2+

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:d^2c^4-12pd^3c^2+8d^2ec^2+16p^2c^4e+256p^2e^3-48d^2e^2+d^2c^4)/(16d^2)

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which is biquadratic and can be solved using square roots.

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Solving for x in terms of y entails solving a quadratic equation, also via square roots.

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Hence we have a solution in terms of square roots and a root of a cubic polynomial.

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This can result in four y's and consequently in eight x's;

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the four roots of (1) can then be determined by trial and error.

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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 whose roots may be variously described as a discrete Fourier transform or a Hadamard matrix transform of the roots.

Related Topics:
Symmetric group - Klein four-group - Normal subgroup - Resolvent - Hadamard matrix

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Suppose ri for i from 0 to 3 are roots of

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

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If we now set

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:s_0 = (r_0 + r_1 + r_2 + r_3)/2

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:s_1 = (r_0 - r_1 + r_2 - r_3)/2

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:s_2 = (r_0 + r_1 - r_2 - r_3)/2

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:s_3 = (r_0 - r_1 - r_2 + r_3)/2

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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 out the polynomial

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:(z^2 - s_1^2)(z^2-s_2^2)(z_3-s_3^2)qquad (2)

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

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

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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.

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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

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F_1 = {x}^{2}+wx+1/2,{w}^{2}+1/2,c-1/2,{rac {{c}^{2}w}{d}}-1/2,{rac {{w}^{5}}{d}}-{rac {c{w}^{3}}{d}}+2,{rac {ew}{d}}

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F_2 =

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then

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

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

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See also

• Linear equation
• Quadratic equation
• Cubic equation
• Quintic equation
• Polynomial
• Lodovico Ferrari
• Girolamo Cardano

Reference

• Ferrari's achievement
• Quartic formula as four single equations