Solution - Finding the roots of polynomials

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((((x4)+(2•(x3)))+23x2)+32x)-128  = 0 
 Step  2  :

Equation at the end of step  2  :

  ((((x4) +  2x3) +  23x2) +  32x) -  128  = 0 

Step  3  :

Polynomial Roots Calculator :

 3.1    Find roots (zeroes) of :       F(x) = x⁴+2x³+8x²+32x-128
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  -128.

 The factor(s) are:

of the Leading Coefficient :  1
 of the Trailing Constant :  1 ,2 ,4 ,8 ,16 ,32 ,64 ,128

 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+2x³+8x²+32x-128 
can be divided by 2 different polynomials,including by  x-2 

Polynomial Long Division :

 3.2    Polynomial Long Division
Dividing :  x⁴+2x³+8x²+32x-128 
                              ("Dividend")
By         :    x-2    ("Divisor")

Quotient :  x³+4x²+16x+64  Remainder:  0 

Polynomial Roots Calculator :

 3.3    Find roots (zeroes) of :       F(x) = x³+4x²+16x+64

     See theory in step 3.1
In this case, the Leading Coefficient is  1  and the Trailing Constant is  64.

 The factor(s) are:

of the Leading Coefficient :  1
 of the Trailing Constant :  1 ,2 ,4 ,8 ,16 ,32 ,64

 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+4x²+16x+64 
can be divided with  x+4 

Polynomial Long Division :

 3.4    Polynomial Long Division
Dividing :  x³+4x²+16x+64 
                              ("Dividend")
By         :    x+4    ("Divisor")

Quotient :  x²+16  Remainder:  0 

Polynomial Roots Calculator :

 3.5    Find roots (zeroes) of :       F(x) = x²+16

     See theory in step 3.1
In this case, the Leading Coefficient is  1  and the Trailing Constant is  16.

 The factor(s) are:

of the Leading Coefficient :  1
 of the Trailing Constant :  1 ,2 ,4 ,8 ,16

 Let us test ....

Polynomial Roots Calculator found no rational roots

Equation at the end of step  3  :

  (x2 + 16) • (x + 4) • (x - 2)  = 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+16 = 0 

 Subtract  16  from both sides of the equation : 
                      x² = -16
 
 When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:  
                      x  =  ± √ -16  

 In Math,  i  is called the imaginary unit. It satisfies   i²  =-1. Both   i   and   -i   are the square roots of   -1 

Accordingly,  √ -16  =
                    √ -1• 16   =
                    √ -1 •√  16   =
                    i •  √ 16

Can  √ 16 be simplified ?

Yes!   The prime factorization of  16   is
   2•2•2•2 
To be able to remove something from under the radical, there have to be  2  instances of it (because we are taking a square i.e. second root).

√ 16   =  √ 2•2•2•2   =2•2•√ 1   =
                ±  4 • √ 1   =
                ±  4

The equation has no real solutions. It has 2 imaginary, or complex solutions.

                      x=  0.0000 + 4.0000 i 
                      x=  0.0000 - 4.0000 i 

Solving a Single Variable Equation :

 4.3      Solve  :    x+4 = 0 

 Subtract  4  from both sides of the equation : 
                      x = -4

Solving a Single Variable Equation :

 4.4      Solve  :    x-2 = 0 

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

Four solutions were found :

1.  x = 2
2.  x = -4
3.   x=  0.0000 - 4.0000 i 
   
4.   x=  0.0000 + 4.0000 i