Solution - Finding the roots of polynomials

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((((x4)-(5•(x3)))+7x2)-5x)+6  = 0 
 Step  2  :

Equation at the end of step  2  :

  ((((x4) -  5x3) +  7x2) -  5x) +  6  = 0 

Step  3  :

Polynomial Roots Calculator :

 3.1    Find roots (zeroes) of :       F(x) = x⁴-5x³+7x²-5x+6
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  6.

 The factor(s) are:

of the Leading Coefficient :  1
 of the Trailing Constant :  1 ,2 ,3 ,6

 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
   x⁴-5x³+7x²-5x+6 
can be divided by 2 different polynomials,including by  x-3 

Polynomial Long Division :

 3.2    Polynomial Long Division
Dividing :  x⁴-5x³+7x²-5x+6 
                              ("Dividend")
By         :    x-3    ("Divisor")

Quotient :  x³-2x²+x-2  Remainder:  0 

Polynomial Roots Calculator :

 3.3    Find roots (zeroes) of :       F(x) = x³-2x²+x-2

     See theory in step 3.1
In this case, the Leading Coefficient is  1  and the Trailing Constant is  -2.

 The factor(s) are:

of the Leading Coefficient :  1
 of the Trailing Constant :  1 ,2

 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
   x³-2x²+x-2 
can be divided with  x-2 

Polynomial Long Division :

 3.4    Polynomial Long Division
Dividing :  x³-2x²+x-2 
                              ("Dividend")
By         :    x-2    ("Divisor")

Quotient :  x²+1  Remainder:  0 

Polynomial Roots Calculator :

 3.5    Find roots (zeroes) of :       F(x) = x²+1

     See theory in step 3.1
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  :

  (x2 + 1) • (x - 2) • (x - 3)  = 0 

Step  4  :

Theory - Roots of a product :

 4.1    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 :

 4.2      Solve  :    x²+1 = 0 

 Subtract  1  from both sides of the equation : 
                      x² = -1
 
 When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:  
                      x  =  ± √ -1  

 In Math,  i  is called the imaginary unit. It satisfies   i²  =-1. Both   i   and   -i   are the square roots of   -1 
The equation has no real solutions. It has 2 imaginary, or complex solutions.

                      x=  0.0000 + 1.0000 i 
                      x=  0.0000 - 1.0000 i 

Solving a Single Variable Equation :

 4.3      Solve  :    x-2 = 0 

 Add  2  to both sides of the equation : 
                      x = 2

Solving a Single Variable Equation :

 4.4      Solve  :    x-3 = 0 

 Add  3  to both sides of the equation : 
                      x = 3

Four solutions were found :

1.  x = 3
2.  x = 2
3.   x=  0.0000 - 1.0000 i 
   
4.   x=  0.0000 + 1.0000 i