Solution - Equations reducible to quadratic form

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "^-20" was replaced by "^(-20)".

Step by step solution :

Step  1  :

Step  2  :

Pulling out like terms :

 2.1     Pull out like factors :

   x - x^(-20)  =   x^(-20) • (x²¹ - 1) 

Trying to factor as a Difference of Cubes:

 2.2      Factoring:  x²¹ - 1 

Theory : A difference of two perfect cubes,  a³ - b³ can be factored into
              (a-b) • (a² +ab +b²)

Proof :  (a-b)•(a²+ab+b²) =
            a³+a²b+ab²-ba²-b²a-b³ =
            a³+(a²b-ba²)+(ab²-b²a)-b³ =
            a³+0+0-b³ =
            a³-b³

Check :  1  is the cube of   1 
Check :  x²¹ is the cube of   x⁷

Factorization is :
             (x⁷ - 1)  •  (x¹⁴ + x⁷ + 1) 

Polynomial Roots Calculator :

 2.3    Find roots (zeroes) of :       F(x) = x⁷ - 1
Polynomial Roots Calculator is a set of methods aimed at finding values of  x  for which   F(x)=0  

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers  x  which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number  P/Q   then  P  is a factor of the Trailing Constant and  Q  is a factor of the Leading Coefficient

In this case, the Leading Coefficient is  1  and the Trailing Constant is  -1.

 The factor(s) are:

of the Leading Coefficient :  1
 of the Trailing Constant :  1

 Let us test ....

The Factor Theorem states that if P/Q is root of a polynomial then this polynomial can be divided by q*x-p Note that q and p originate from P/Q reduced to its lowest terms

In our case this means that
   x⁷ - 1 
can be divided with  x - 1 

Polynomial Long Division :

 2.4    Polynomial Long Division
Dividing :  x⁷ - 1 
                              ("Dividend")
By         :    x - 1    ("Divisor")

Quotient :  x⁶+x⁵+x⁴+x³+x²+x+1  Remainder:  0 

Polynomial Roots Calculator :

 2.5    Find roots (zeroes) of :       F(x) = x⁶+x⁵+x⁴+x³+x²+x+1

     See theory in step 2.3
In this case, the Leading Coefficient is  1  and the Trailing Constant is  1.

 The factor(s) are:

of the Leading Coefficient :  1
 of the Trailing Constant :  1

 Let us test ....

Polynomial Roots Calculator found no rational roots

Trying to factor by splitting the middle term

 2.6     Factoring  x¹⁴+x⁷+1 

The first term is,  x¹⁴  its coefficient is  1 .
The middle term is,  +x⁷  its coefficient is  1 .
The last term, "the constant", is  +1 

Step-1 : Multiply the coefficient of the first term by the constant   1 • 1 = 1 

Step-2 : Find two factors of  1  whose sum equals the coefficient of the middle term, which is   1 .

Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored

Equation at the end of step  2  :

  x(-20)•(x6+x5+x4+x3+x2+x+1)•(x-1)•(x14+x7+1)  = 0 

Step  3  :

Theory - Roots of a product :

 3.1    A product of several terms equals zero. 

 When a product of two or more terms equals zero, then at least one of the terms must be zero. 

 We shall now solve each term = 0 separately 

 In other words, we are going to solve as many equations as there are terms in the product 

 Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

 3.2      Solve  :    x^(-20) = 0 

  This equation has no solution !!

We actually looking at 1/

Equations of order 5 or higher :

 3.3     Solve   x⁶+x⁵+x⁴+x³+x²+x+1 = 0

In search of an interavl at which the above polynomial changes sign, from negative to positive or the other wayaround.

Method of search: Calculate polynomial values for all integer points between x=-20 and x=+20

No interval at which a change of sign occures has been found. Consequently, Bisection Approximation can not be used. As this is a polynomial of an even degree it may not even have any real (as opposed to imaginary) roots

Solving a Single Variable Equation :

 3.4      Solve  :    x-1 = 0 

 Add  1  to both sides of the equation : 
                      x = 1

Solving a Single Variable Equation :

Equations which are reducible to quadratic :

 3.5     Solve   x¹⁴+x⁷+1 = 0

This equation is reducible to quadratic. What this means is that using a new variable, we can rewrite this equation as a quadratic equation Using  w , such that  w = x⁷  transforms the equation into :
 w²+w+1 = 0

Solving this new equation using the quadratic formula we get two imaginary solutions :
   w = -0.5000 ± 0.8660 i 
Now that we know the value(s) of  w , we can calculate  x  since  x  is the seventh root of   w  

Since we are speaking 7th root, each of the two imaginary solutions of has 7 roots

Tiger finds these roots using de Moivre's Formula

The 7th roots of  -0.500 + 0.866 i   are:

  x =  0.956 + 0.295 i 
  x =  0.365 + 0.931 i 
  x = -0.500 + 0.866 i 
  x = -0.989 + 0.149 i 
  x = -0.733 -0.680 i 
  x =  0.075 -0.997 i 
  x =  0.826 -0.563 i 

7th roots of  -0.500- 0.866 i  :
  x =  0.826  + 0.563 i  
  x =  0.075  + 0.997 i  
  x = -0.733  + 0.680 i  
  x = -0.989 - 0.149 i 
  x = -0.500 - 0.866 i 
  x =  0.365 - 0.931 i 
  x =  0.956 - 0.295 i 
 

15 solutions were found :

1.   x = 0.956 - 0.295 i
2.   x = 0.365 - 0.931 i
3.   x = -0.500 - 0.866 i
4.   x = -0.989 - 0.149 i
5.   x = -0.733 + 0.680 i
6.   x = 0.075 + 0.997 i
7.   x = 0.826 + 0.563 i
8.   x = 0.826 -0.563 i
9.   x = 0.075 -0.997 i
10.   x = -0.733 -0.680 i
11.   x = -0.989 + 0.149 i
12.   x = -0.500 + 0.866 i
13.   x = 0.365 + 0.931 i
14.   x = 0.956 + 0.295 i
15.  x = 1