Solution - Factoring binomials using the difference of squares

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((2•3x25) • x) -  6  = 0 
 Step  2  :
 Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   6x²⁶ - 6  =   6 • (x²⁶ - 1) 

Trying to factor as a Difference of Squares :

 3.2      Factoring:  x²⁶ - 1 

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 : 1 is the square of 1
Check :  x²⁶  is the square of  x¹³ 

Factorization is :       (x¹³ + 1)  •  (x¹³ - 1) 

Equation at the end of step  3  :

  6 • (x13 + 1) • (x13 - 1)  = 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.

Equations which are never true :

 4.2      Solve :    6   =  0

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

Solving a Single Variable Equation :

 4.3      Solve  :    x¹³+1 = 0 

 Subtract  1  from both sides of the equation : 
                      x¹³ = -1
                     x  =  13th root of (-1) 

 Negative numbers have real 13th roots.
 13th root of (-1) = ¹³√ -1• 1  = ¹³√ -1 • ¹³√ 1  =(-1)•¹³√ 1 

The equation has one real solution, a negative number This solution is  x = negative

Solving a Single Variable Equation :

 4.4      Solve  :    x¹³-1 = 0 

 Add  1  to both sides of the equation : 
                      x¹³ = 1
                     x  =  13th root of (1) 

 The equation has one real solution
This solution is  x =

Two solutions were found :

1.  x =
2.  x = negative