Solution - Simplification or other simple results

Step  1  :

Equation at the end of step  1  :

  (((1•(x-1))•((x2)-2))+((2•(x+1))•((x2)-1)))-((x+5)•(x+1)•(x-1))

Step  2  :

Equation at the end of step  2  :

  (((1•(x-1))•((x2)-2))+((2•(x+1))•((x2)-1)))-(x+5)•(x+1)•(x-1)

Step  3  :

Equation at the end of step  3  :

  (((1•(x-1))•((x2)-2))+(2•(x+1)•(x2-1)))-(x+5)•(x+1)•(x-1)

Step  4  :

Trying to factor as a Difference of Squares :

 4.1      Factoring:  x²-1 

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 : 1 is the square of 1
Check :  x²  is the square of  x¹ 

Factorization is :       (x + 1)  •  (x - 1) 

Multiplying Exponential Expressions :

 4.2    Multiply  (x + 1)  by  (x + 1) 

The rule says : To multiply exponential expressions which have the same base, add up their exponents.

In our case, the common base is  (x+1)  and the exponents are :
          1 , as  (x+1)  is the same number as  (x+1)¹ 
 and   1 , as  (x+1)  is the same number as  (x+1)¹ 
The product is therefore,  (x+1)⁽¹⁺¹⁾ = (x+1)² 

Equation at the end of step  4  :

  (((1•(x-1))•((x2)-2))+2•(x+1)2•(x-1))-(x+5)•(x+1)•(x-1)

Step  5  :

Equation at the end of step  5  :

  (((x-1)•(x2-2))+2•(x+1)2•(x-1))-(x+5)•(x+1)•(x-1)

Step  6  :

Evaluate an expression :

 6.1      Factoring:  x²-2 

Check : 2 is not a square !!

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

Equation at the end of step  6  :

  ((x-1)•(x2-2)+2•(x+1)2•(x-1))-(x+5)•(x+1)•(x-1)
 Step  7  :

Pulling out like terms :

 7.1      Pull out     x-1 

After pulling out, we are left with :
      (x-1) • ( 1  *  (3x²+4x) +( (x+5)  *  (x+1)  *  (-1) ))

Trying to factor by splitting the middle term

 7.2     Factoring  2x²-2x-5 

The first term is,  2x²  its coefficient is  2 .
The middle term is,  -2x  its coefficient is  -2 .
The last term, "the constant", is  -5 

Step-1 : Multiply the coefficient of the first term by the constant   2 • -5 = -10 

Step-2 : Find two factors of  -10  whose sum equals the coefficient of the middle term, which is   -2 .

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

Final result :

  (x - 1) • (2x2 - 2x - 5)