Solution - Finding the roots of polynomials

Step by step solution :

Step  1  :

             1
 Simplify   ——
            x5

Equation at the end of step  1  :

         1     
  (0 -  ——) -  1  = 0 
        x5     

Step  2  :

Rewriting the whole as an Equivalent Fraction :

 2.1   Subtracting a whole from a fraction

Rewrite the whole as a fraction using  x⁵  as the denominator :

         1     1 • x5
    1 =  —  =  ——————
         1       x5  

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

 2.2       Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

 -1 - (x5)     -x5 - 1
 —————————  =  ———————
    x5           x5   

Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   -x⁵ - 1  =   -1 • (x⁵ + 1) 

Polynomial Roots Calculator :

 3.2    Find roots (zeroes) of :       F(x) = x⁵ + 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
   x⁵ + 1 
can be divided with  x + 1 

Polynomial Long Division :

 3.3    Polynomial Long Division
Dividing :  x⁵ + 1 
                              ("Dividend")
By         :    x + 1    ("Divisor")

Quotient :  x⁴-x³+x²-x+1  Remainder:  0 

Polynomial Roots Calculator :

 3.4    Find roots (zeroes) of :       F(x) = x⁴-x³+x²-x+1

     See theory in step 3.2
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

Equation at the end of step  3  :

  (-x4 + x3 - x2 + x - 1) • (x + 1)
  —————————————————————————————————  = 0 
                 x5                

Step  4  :

When a fraction equals zero :

 4.1    When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

  (-x4+x3-x2+x-1)•(x+1)
  ————————————————————— • x5 = 0 • x5
           x5          

Now, on the left hand side, the  x⁵  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   (-x⁴+x³-x²+x-1)  •  (x+1)  = 0

Theory - Roots of a product :

 4.2    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.

Quartic Equations :

 4.3     Solve   -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 :

 4.4      Solve  :    x+1 = 0 

 Subtract  1  from both sides of the equation : 
                      x = -1

One solution was found :