Solution - Simplifying radicals

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  (((2 • (x3)) -  (2•32x2)) +  8x) -  72  = 0 
 Step  2  :

Equation at the end of step  2  :

  ((2x3 -  (2•32x2)) +  8x) -  72  = 0 

Step  3  :

Step  4  :

Pulling out like terms :

 4.1     Pull out like factors :

   2x³ - 18x² + 8x - 72  = 

  2 • (x³ - 9x² + 4x - 36) 

Checking for a perfect cube :

 4.2    x³ - 9x² + 4x - 36  is not a perfect cube

Trying to factor by pulling out :

 4.3      Factoring:  x³ - 9x² + 4x - 36 

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1:  4x - 36 
Group 2:  -9x² + x³ 

Pull out from each group separately :

Group 1:   (x - 9) • (4)
Group 2:   (x - 9) • (x²)
               -------------------
Add up the two groups :
               (x - 9)  •  (x² + 4) 
Which is the desired factorization

Polynomial Roots Calculator :

 4.4    Find roots (zeroes) of :       F(x) = x² + 4
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  4.

 The factor(s) are:

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

 Let us test ....

Polynomial Roots Calculator found no rational roots

Equation at the end of step  4  :

  2 • (x2 + 4) • (x - 9)  = 0 

Step  5  :

Theory - Roots of a product :

 5.1    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.

Equations which are never true :

 5.2      Solve :    2   =  0

This equation has no solution.
A a non-zero constant never equals zero.

Solving a Single Variable Equation :

 5.3      Solve  :    x²+4 = 0 

 Subtract  4  from both sides of the equation : 
                      x² = -4
 
 When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:  
                      x  =  ± √ -4  

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

Accordingly,  √ -4  =
                    √ -1• 4   =
                    √ -1 •√  4   =
                    i •  √ 4

Can  √ 4 be simplified ?

Yes!   The prime factorization of  4   is
   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).

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

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

                      x=  0.0000 + 2.0000 i 
                      x=  0.0000 - 2.0000 i 

Solving a Single Variable Equation :

 5.4      Solve  :    x-9 = 0 

 Add  9  to both sides of the equation : 
                      x = 9

Three solutions were found :

1.  x = 9
2.   x=  0.0000 - 2.0000 i 
   
3.   x=  0.0000 + 2.0000 i