Solution - Factoring binomials using the difference of squares

Step  1  :

Equation at the end of step  1  :

  (((3 • (x4)) +  5x3) +  81x) +  135
 Step  2  :

Equation at the end of step  2  :

  ((3x4 +  5x3) +  81x) +  135

Step  3  :

Checking for a perfect cube :

 3.1    3x⁴+5x³+81x+135  is not a perfect cube

Trying to factor by pulling out :

 3.2      Factoring:  3x⁴+5x³+81x+135 

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

Group 1:  81x+135 
Group 2:  3x⁴+5x³ 

Pull out from each group separately :

Group 1:   (3x+5) • (27)
Group 2:   (3x+5) • (x³)
               -------------------
Add up the two groups :
               (3x+5)  •  (x³+27) 
Which is the desired factorization

Trying to factor as a Sum of Cubes :

 3.3      Factoring:  x³+27 

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 :  27  is the cube of   3 
Check :  x³ is the cube of   x¹

Factorization is :
             (x + 3)  •  (x² - 3x + 9) 

Trying to factor by splitting the middle term

 3.4     Factoring  x² - 3x + 9 

The first term is,  x²  its coefficient is  1 .
The middle term is,  -3x  its coefficient is  -3 .
The last term, "the constant", is  +9 

Step-1 : Multiply the coefficient of the first term by the constant   1 • 9 = 9 

Step-2 : Find two factors of  9  whose sum equals the coefficient of the middle term, which is   -3 .

Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored

Final result :

  (x + 3) • (x2 - 3x + 9) • (3x + 5)