Solution - Factoring binomials using the difference of squares

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  (3x29 • x) -  12  = 0 
 Step  2  :
 Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   3x³⁰ - 12  =   3 • (x³⁰ - 4) 

Trying to factor as a Difference of Squares :

 3.2      Factoring:  x³⁰ - 4 

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 : 4 is the square of 2
Check :  x³⁰  is the square of  x¹⁵ 

Factorization is :       (x¹⁵ + 2)  •  (x¹⁵ - 2) 

Trying to factor as a Sum of Cubes :

 3.3      Factoring:  x¹⁵ + 2 

Theory : A sum of two perfect cubes,  a³ + b³ can be factored into  :
             (a+b) • (a²-ab+b²)
Proof  : (a+b) • (a²-ab+b²) =
    a³-a²b+ab²+ba²-b²a+b³ =
    a³+(a²b-ba²)+(ab²-b²a)+b³=
    a³+0+0+b³=
    a³+b³

Check :  2  is not a cube !!
Ruling : Binomial can not be factored as the difference of two perfect cubes

Trying to factor as a Difference of Cubes:

 3.4      Factoring:  x¹⁵ - 2 

Theory : A difference of two perfect cubes,  a³ - b³ can be factored into
              (a-b) • (a² +ab +b²)

Proof :  (a-b)•(a²+ab+b²) =
            a³+a²b+ab²-ba²-b²a-b³ =
            a³+(a²b-ba²)+(ab²-b²a)-b³ =
            a³+0+0-b³ =
            a³-b³

Check :  2  is not a cube !!
Ruling : Binomial can not be factored as the difference of two perfect cubes

Equation at the end of step  3  :

  3 • (x15 + 2) • (x15 - 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.

Equations which are never true :

 4.2      Solve :    3   =  0

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

Solving a Single Variable Equation :

 4.3      Solve  :    x¹⁵+2 = 0 

 Subtract  2  from both sides of the equation : 
                      x¹⁵ = -2
                     x  =  15th root of (-2) 

 Negative numbers have real 15th roots.
 15th root of (-2) = ¹⁵√ -1• 2  = ¹⁵√ -1 • ¹⁵√ 2  =(-1)•¹⁵√ 2 

The equation has one real solution, a negative number This solution is  x = negative 15th root of 2 = -1.0473

Solving a Single Variable Equation :

 4.4      Solve  :    x¹⁵-2 = 0 

 Add  2  to both sides of the equation : 
                      x¹⁵ = 2
                     x  =  15th root of (2) 

 The equation has one real solution
This solution is  x = 15th root of 2 = 1.0473

Two solutions were found :

1.  x = 15th root of 2 = 1.0473
2.  x = negative 15th root of 2 = -1.0473