Solution - Factoring binomials using the difference of squares

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  (((a3) -  11a2) -  9a) +  99  = 0 

Step  2  :

Checking for a perfect cube :

 2.1    a³-11a²-9a+99  is not a perfect cube

Trying to factor by pulling out :

 2.2      Factoring:  a³-11a²-9a+99 

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1:  -9a+99 
Group 2:  a³-11a² 

Pull out from each group separately :

Group 1:   (a-11) • (-9)
Group 2:   (a-11) • (a²)
               -------------------
Add up the two groups :
               (a-11)  •  (a²-9) 
Which is the desired factorization

Trying to factor as a Difference of Squares :

 2.3      Factoring:  a²-9 

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 : 9 is the square of 3
Check :  a²  is the square of  a¹ 

Factorization is :       (a + 3)  •  (a - 3) 

Equation at the end of step  2  :

  (a + 3) • (a - 3) • (a - 11)  = 0 

Step  3  :

Theory - Roots of a product :

 3.1    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 :

 3.2      Solve  :    a+3 = 0 

 Subtract  3  from both sides of the equation : 
                      a = -3

Solving a Single Variable Equation :

 3.3      Solve  :    a-3 = 0 

 Add  3  to both sides of the equation : 
                      a = 3

Solving a Single Variable Equation :

 3.4      Solve  :    a-11 = 0 

 Add  11  to both sides of the equation : 
                      a = 11

Three solutions were found :

1.  a = 11
2.  a = 3
3.  a = -3