Solution - Factoring binomials using the difference of squares

Reformatting the input :

Changes made to your input should not affect the solution:

 (1): "a5"   was replaced by   "a^5". 

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  (a2) -  (2•5a52)  = 0 
 Step  2  :
 Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   a² - 10a⁵²  =   -a² • (10a⁵⁰ - 1) 

Trying to factor as a Difference of Squares :

 3.2      Factoring:  10a⁵⁰ - 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 :  10  is not a square !!

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

Equation at the end of step  3  :

  -a2 • (10a50 - 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  :    -a² = 0 

 Multiply both sides of the equation by (-1) :  a² = 0


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

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

Solving a Single Variable Equation :

 4.3      Solve  :    10a⁵⁰-1 = 0 

 Add  1  to both sides of the equation : 
                      10a⁵⁰ = 1
Divide both sides of the equation by 10:
                     a⁵⁰ = 1/10 = 0.100
                     a  =  50th root of (1/10) 

 The equation has two real solutions  
 These solutions are  a = 50th root of ( 0.100) = ± 0.95499  
 

Three solutions were found :

1.  a = 50th root of ( 0.100) = ± 0.95499
2.  a = 0