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Solution - Factoring binomials using the difference of squares

x=±root[10]2=±1.0718
x=±root[10]{2}=±1.0718
x=0
x=0

Step by Step Solution

Reformatting the input :

Changes made to your input should not affect the solution:

 (1): "x1"   was replaced by   "x^1". 

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

                     2*(x^6)-(x^16)=0 

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  2x6 -  x16  = 0 

Step  2  :

Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   2x6 - x16  =   -x6 • (x10 - 2) 

Trying to factor as a Difference of Squares :

 3.2      Factoring:  x10 - 2 

Theory : A difference of two perfect squares,  A2 - B2  can be factored into  (A+B) • (A-B)

Proof :  (A+B) • (A-B) =
         A2 - AB + BA - B2 =
         A2 - AB + AB - B2 =
         A2 - B2

Note :  AB = BA is the commutative property of multiplication.

Note :  - AB + AB equals zero and is therefore eliminated from the expression.

Check : 2 is not a square !!

Ruling : Binomial can not be factored as the difference of two perfect squares.

Equation at the end of step  3  :

  -x6 • (x10 - 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  :    -x6 = 0 

 
Multiply both sides of the equation by (-1) :  x6 = 0


                     x  =  6th root of (0) 

 
Any root of zero is zero. This equation has one solution which is  x = 0

Solving a Single Variable Equation :

 4.3      Solve  :    x10-2 = 0 

 
Add  2  to both sides of the equation : 
 
                     x10 = 2
                     x  =  10th root of (2) 

 
The equation has two real solutions  
 
These solutions are  x = ± 10th root of 2 = ± 1.0718  
 

Three solutions were found :

  1.  x = ± 10th root of 2 = ± 1.0718
  2.  x = 0

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