Solution - Factoring binomials using the difference of squares

Reformatting the input :

Changes made to your input should not affect the solution:

 (1): "x5"   was replaced by   "x^5". 

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  (x2) -  (3•5x50)  = 0 
 Step  2  :
 Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   x² - 15x⁵⁰  =   -x² • (15x⁴⁸ - 1) 

Trying to factor as a Difference of Squares :

 3.2      Factoring:  15x⁴⁸ - 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 :  15  is not a square !!

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

Trying to factor as a Difference of Cubes:

 3.3      Factoring:  15x⁴⁸ - 1 

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 :  15  is not a cube !!

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

Equation at the end of step  3  :

  -x2 • (15x48 - 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.

Solving a Single Variable Equation :

 4.2      Solve  :    -x² = 0 

 Multiply both sides of the equation by (-1) :  x² = 0


 
 When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:  
                      x  =  ± √ 0  

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

Solving a Single Variable Equation :

 4.3      Solve  :    15x⁴⁸-1 = 0 

 Add  1  to both sides of the equation : 
                      15x⁴⁸ = 1
Divide both sides of the equation by 15:
                     x⁴⁸ = 1/15 = 0.067
                     x  =  48th root of (1/15) 

 The equation has two real solutions  
 These solutions are  x = 48th root of ( 0.067) = ± 0.94514  
 

Three solutions were found :

1.  x = 48th root of ( 0.067) = ± 0.94514
2.  x = 0