Home #1 Mathematics The Formulae to find the roots of polynomials up to 4th degree
The Formulae to find the roots of polynomials up to 4th degree PDF Print E-mail
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Wednesday, 13 August 2008 00:00

May be you know, may be you don't know, but only the polynomials which degree is lower or equal than 4, would have a solution using radicals.

Problems like the duplication of the cube or the trisection of an angle are connected with irreducible polynomials.

You know that the solution of the equation:

a x + b = c, (1)

is

x = (c - b)/a

The polynomial (1), is an equation of polynomial of degree #1.

If you have a polynomial of degree (2):

a x2+ b x + c = 0,

Then,

a2 x2+ b a x + c a = 0,

and,

(2 a x)2+ 4 b a x + 4 a c = 0,

(2 a x)2+ 4 b a x + b2 + 4 a c = b2,

So,

(2ax + b)2= b2- 4 a c,

and therefore,

x = (-b ± √(b2- 4 a c) )/ 2 a

 

Now, we will go to solve the polynomial of the the third degree. Because the first coefficient of a polynomial will be different than zero, we can divide for it, so the coefficient of the highest degree will be 1.

In other words, we will star for simplicity, using:

x3+ a x2 + b x + c = 0,

then, we can reduce this equation to,

y3+ p y + q = 0,

using the variable change: x = y - 1/3 a.

In fact,

(y - 1/3 a)3+ a (y - 1/3a)2 + b (y - 1/3a) + c = 0,

means that,

y3 - 3 y2 (1/3) a + 3 y (1/3)2 a2 - 1/9 a3 + a (y2-2/3 a y + 1/9 a) + b y - ba/3 + c = 0,

In other words,

y3+ p y + q = 0, (2)

where,

p = (1/3 a2-2/3 a + b) and q = (c-1/9a3-ba/3)

So, is sufficient to solve (2), without the member with the quadratic (2) power.

Now, we apply the substitution: y = u + v. Then, y, will be a root u3+v3= -b, and u v = -1/3 a. In other words, u3 is a root of the quadratic of the equation in u3,

(u3)2+ b (u3) - a3/27 = 0, where uv = -1/3 a.

Now, if

u1=3√ (1/2 (-b + √(b2+4a3/27) )), and v1=-1/3 a/u1 and w = -1/2 + 1/2 i √3,

then the three roots are:

z1= u1+v,

z2= w u1 + w2 v1 and

z3= w2 u1+ w v1

Had been proved that there are always a real root. This formula was published by Girolamo Cardano (1501-1576) after Niccolò Fontana (1500-1557), Tartaglia show to him. Cardano give credits to Tartaglia, in him publication.

At last, if we want,

x4+ a3 x3 + a2x2 + a1x + a0 = 0,

then its root can be found solving the two quadratic equations:

x2+ (A + C) x + (B + D) = 0

x2+ (A - C) x + (B - D) = 0,

where,

A = a3/2, B = y0/2 and

D = √ B2- a0 and

C = √A2-a2+y0 if D = 0, and C= (AB - a1/2)/D, othwerwise,

y0 is the largest real root of the cubic below:

y3+b2y2+b1y+b0=0

where,

b2= - a2,

b1= a3a1-4a0,

b0=a0 (4a2 - a32) - a12

Thanks,

Giovanni A. Orlando.

 

 

 

 

 

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Last Updated on Wednesday, 13 August 2008 16:47