Solution - Factoring multivariable polynomials

Step  1  :

Equation at the end of step  1  :

  ((8•(x3))-(125•(y3)))   (((4•(x2))+10xy)+(25•(y2)))
  ————————————————————— ÷ ———————————————————————————
       ((2x+5y)3)           (((4•(x2))+20xy)+52y2)   
 Step  2  :

Equation at the end of step  2  :

  ((8•(x3))-(125•(y3)))   (((4•(x2))+10xy)+(25•(y2)))
  ————————————————————— ÷ ———————————————————————————
       ((2x+5y)3)             ((22x2+20xy)+52y2)     
 Step  3  :

Equation at the end of step  3  :

  ((8•(x3))-(125•(y3)))   (((4•(x2))+10xy)+52y2)
  ————————————————————— ÷ ——————————————————————
       ((2x+5y)3)            (4x2+20xy+25y2)    
 Step  4  :

Equation at the end of step  4  :

  ((8•(x3))-(125•(y3)))   ((22x2+10xy)+52y2)
  ————————————————————— ÷ ——————————————————
       ((2x+5y)3)          (4x2+20xy+25y2)  

Step  5  :

            4x2 + 10xy + 25y2
 Simplify   —————————————————
            4x2 + 20xy + 25y2

Trying to factor a multi variable polynomial :

 5.1    Factoring    4x² + 10xy + 25y² 

Try to factor this multi-variable trinomial using trial and error 

 Factorization fails

Trying to factor a multi variable polynomial :

 5.2    Factoring    4x² + 20xy + 25y² 

Try to factor this multi-variable trinomial using trial and error 

 Found a factorization  :  (2x + 5y)•(2x + 5y)

Detecting a perfect square :

 5.3    4x²  +20xy  +25y²  is a perfect square 

 It factors into  (2x+5y)•(2x+5y)
which is another way of writing  (2x+5y)²

How to recognize a perfect square trinomial:  

 • It has three terms  

 • Two of its terms are perfect squares themselves  

 • The remaining term is twice the product of the square roots of the other two terms

Equation at the end of step  5  :

  ((8•(x3))-(125•(y3)))   (4x2+10xy+25y2)
  ————————————————————— ÷ ———————————————
       ((2x+5y)3)            (2x+5y)2    
 Step  6  :

Equation at the end of step  6  :

  ((8•(x3))-53y3)   (4x2+10xy+25y2)
  ——————————————— ÷ ———————————————
     (2x+5y)3          (2x+5y)2    
 Step  7  :

Equation at the end of step  7  :

  (23x3 - 53y3)    (4x2 + 10xy + 25y2)
  ————————————— ÷ ———————————————————
   (2x + 5y)3         (2x + 5y)2     

Step  8  :

            8x3 - 125y3
 Simplify   ———————————
            (2x + 5y)3 

Trying to factor as a Difference of Cubes:

 8.1      Factoring:  8x³ - 125y³ 

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 :  8  is the cube of  2 

Check :  125  is the cube of   5 
Check :  x³ is the cube of   x¹

Check :  y³ is the cube of   y¹

Factorization is :
             (2x - 5y)  •  (4x² + 10xy + 25y²) 

Trying to factor a multi variable polynomial :

 8.2    Factoring    4x² + 10xy + 25y² 

Try to factor this multi-variable trinomial using trial and error 

 Factorization fails

Equation at the end of step  8  :

  (2x - 5y) • (4x2 + 10xy + 25y2)   (4x2 + 10xy + 25y2)
  ——————————————————————————————— ÷ ———————————————————
            (2x + 5y)3                  (2x + 5y)2     

Step  9  :

         (2x-5y)•(4x2+10xy+25y2)      4x2+10xy+25y2
 Divide  ———————————————————————  by  —————————————
                (2x+5y)3                (2x+5y)2   

 9.1    Dividing fractions

To divide fractions, write the divison as multiplication by the reciprocal of the divisor :

(2x - 5y) • (4x2 + 10xy + 25y2)     4x2 + 10xy + 25y2       (2x - 5y) • (4x2 + 10xy + 25y2)          (2x + 5y)2    
———————————————————————————————  ÷  —————————————————   =   ———————————————————————————————  •  ———————————————————
          (2x + 5y)3                   (2x + 5y)2                     (2x + 5y)3                (4x2 + 10xy + 25y2)

Canceling Out :

 9.2    Cancel out  (4x² + 10xy + 25y²)  which appears on both sides of the fraction line.

Dividing Exponential Expressions :

 9.3    Divide  (2x + 5y)²   by  (2x + 5y)³  

The rule says : To divide exponential expressions which have the same base, subtract their exponents.

In our case, the common base is  (2x+5y)  and the exponents are :
          2
 and   3
The quotient is therefore,  (2x+5y)^(2-3) = (2x+5y)^(-1) 

Note that the quotient has a negative exponent.
Rewrite the quotient under the fraction line changing its exponent to be positive: 1/(2x+5y)¹

Omit the '1' in the exponent altogether. Anything to the first power is the number itself so there is usually no reason to write down the '1

Final result :

  2x - 5y
  ———————
  2x + 5y