Elementary theory of cubics and quartics

It is proved that cubic and quartic equations in real numbers can be solved elementarily, avoiding complex numbers and derivatives. Corresponding algorithms are presented.

The school with the third and fourth degree equations encounters rarely (actually only when they have rational solutions).They are not considered in mathematical classes or even in facultative courses -the widespread belief is that they are needed to deal with complex numbers (see [1]).The purpose of this paper -to show that to solve elementary these equations is perfectly possible, and school knowledge is sufficient for this.
Cubic equations.General form of cubic (third degree) equation is The equation can be divided by a = 0, and this form of the equation is called general one too: The equation can be further simplified: entering a new variable y by substitution x = y + k and selecting parameter k one can achieve that the equation has no longer a member of the second degree.Really, after inserting x = y + k in equation (2), only the two first addends x 3 and bx 2 give the terms with y 2 .Because then only terms ky 2 + by 2 + 2ky 2 have y 2 .Thus y 2 disappears if 3k + b = 0, i.e., k = −b/3.Thus, for equation (2) had not a quadratic member, the substitution can be made x = y − b/3.The resulting equation is called reduced one (the variable here again is denoted by x): The number of solutions.Cubic equation ( 3) always has at least one solution: on the left standing cubic function f (x) is positive for sufficiently large x and negative for large negative x, so f (x) graph crosses the x-axis.Cubic equation may have one, two or three solutions -for example, equations have sets of solutions, respectively, {1}, {1, 2}, {1, 2, 3}.
More than three solutions cubic equation can have not.Indeed, suppose that equation (3) has a solution α.This means that α 3 + pα + q = 0. Now (3) it is easy to factorize: Thus, other solutions of equation ( 3) we find from the quadratic equation So equation (4) can add to α 0, 1 or maximum 2 new (i.e., not coinciding with α) solutions of equation (3).We see that the cubic equation solving and determination of number of solutions knowing at least one solution becomes trivial.That is why it is very easy to solve the cubic equation, which has a rational solution.By the way, it is easy to find rational solutions of every equation of any degree with rational coefficients.For example, if the coefficients of an equation are integers and the first coefficient is 1 (such a form we can give to every equation with rational coefficients), then rational solutions can be only integers -positive and negative divisors of the free member (see [2]).So it is worth the solving of every equation with rational coefficients always begin from rational solutions.
Case p> 0. The cubic equation ( 3) can be further simplified to make the modulus of coefficient p = 0 equal to 3 (when p = 0, then equation (3) becomes x 3 = −q and comprises a unique solution x = − 3 √ q).Again, we use a simple linear substitution x = ky and properly select k.Our equation ( 3) is converted into k 3 y 3 + pky + q = 0, y 3 + py/k 2 + q/k 3 = 0.

J.J. Mačys
Let's make sure that equation (5) has no more solutions.Recall equation (4)now it looks like this: The discriminant of the quadratic equation is equal to Since it is negative, equation (5) has no more solutions.Thus, equation (3) in the case p > 0 has the only solution.We find it going back from y to x = y p/3.Case p< 0. It is more difficult to solve equation (3), when p is negative.We apply substitution x = ky again.Now in the equation Denoting the resulting free member −2m, we have the equation We apply substitution y = z + 1/z: Here awaits us a surprise -equation (7) has solutions not always, and we have to consider three subcases: m 2 > 1, m 2 = 1 and m 2 < 1.
Subcase p < 0, m 2 > 1 (the unique solution).If m 2 > 1, then and both values of z give the same solution: Let us convince that this solution is unique.The modulus of this solution is greater than 2: Really, y 2 = 4 would mean y = ±2, and then from equation (6) (but in our case must be m 2 > 1).Other solutions could be given by (4) equation, which now looks like this: The discriminant of it is negative, and equation (4) has no solutions.Thus, equation (6) in case of p < 0, m 2 > 1 has the unique solution.
Subcase p < 0, m 2 < 1 (three solutions).If m 2 < 1, then equation ( 7) has no solutions, and substitution y = z + 1/z did not find a solution.But here helps trigonometry: the substitution y = 2z puts equation ( 6) in the form 8z 3 −6z −2m = 0, i.e. 4z 3 − 3z − m = 0. Do substitution z = cos ϕ: This equation has solutions specifically when |m| < 1 (that is m 2 < 1).Further, Based on the reduction formulas we make sure we get only three different y values (more than three solutions cubic equation can have not!): We return to x.General cubic equation ( 3) is solved.
An example.Consider a concrete equation in the form (3): Solve it anew, "without theory".(Of course, the theory is always born of concrete examples.)Think of the school formulas having something like x 3 − 3x.There comes to mind the formula and its particular case (a ± 1/a) 3 = a 3 ± 1/a 3 ± 3(a ± 1/a).
As a rule namely formulas of this section you find in manuals (cf.[2]), so one can get solutions without solving equation (3).

Quartic equation. Common fourth degree equation is
It can be divided by a, so it is enough to consider the equation After substitution x = y − a/4, we eliminate of it a member of the third degree, so it is enough to know how to solve the equation If the fourth degree equation has rational solutions (at least one rational solution α), it is simple to solve it: the left-hand side can be factorized again, separating the x − α, and it remains to solve the cubic equation.Consider (9) equation (when it does not have solutions; naturally, we can also deal with when it has).
If c = 0, equation ( 9) is biquadrate, and adopting the new variable x 2 = y it becomes quadratic one.By the way, it is easy to decompose biquadrate trinomial If c = 0, then the left side of equation ( 9) is always possible to write as the difference of two squares: Thus the left-hand side can be factorized into two quadratic trinomials, and you solve two quadratic equations.
Left to figure out how to find a suitable m, p and n values.In equation removing brackets and comparing the coefficients of x 2 , x 1 and x 0 , we have the system of equations for suitable m, p, n to find:  After multiplying by 4p, we get the cubic equation (cf.[1,3]) This equation has positive solution (namely, this new observation is the fourth degree equation solving key).the function f (p) on the left at p = 0 is equal to f (0) = −c 2 , namely negative (recall c = 0) and for the sufficiently large positive p it is positive, so the graph of f (p) intersects p axis at any point p 0 ∈ (0, ∞).Since we can solve the cubic equation, we find positive solution p 0 , receiving the solution of the system (10) (m, p, n) = (b/2 + p 0 /2, p 0 , − 1 2 c/p 0 ).Thus, we are able to decompose the left side of equation ( 9) and, consequently, to solve any fourth order equation.It may have 1, 2, 3, 4 or no solutions (e.g.x 4 + 1 = 0 has no solutions).
It turns out that to solve the general equation of the fifth degree is not possible.However, the most important statement (the fundamental theorem of algebra) remains correct [1]: each n-th degree polynomial is a product of linear and quadratic factors.