Solution - Factoring binomials using the difference of squares

Step  1  :

Equation at the end of step  1  :

  (2x23 • x) -  2
 Step  2  :
 Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   2x²⁴ - 2  =   2 • (x²⁴ - 1) 

Trying to factor as a Difference of Squares :

 3.2      Factoring:  x²⁴ - 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 :  x²⁴  is the square of  x¹² 

Factorization is :       (x¹² + 1)  •  (x¹² - 1) 

Trying to factor as a Sum of Cubes :

 3.3      Factoring:  x¹² + 1 

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 :  1  is the cube of   1 
Check :  x¹² is the cube of   x⁴

Factorization is :
             (x⁴ + 1)  •  (x⁸ - x⁴ + 1) 

Polynomial Roots Calculator :

 3.4    Find roots (zeroes) of :       F(x) = x⁴ + 1
Polynomial Roots Calculator is a set of methods aimed at finding values of  x  for which   F(x)=0  

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers  x  which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number  P/Q   then  P  is a factor of the Trailing Constant and  Q  is a factor of the Leading Coefficient

In this case, the Leading Coefficient is  1  and the Trailing Constant is  1.

 The factor(s) are:

of the Leading Coefficient :  1
 of the Trailing Constant :  1

 Let us test ....

Polynomial Roots Calculator found no rational roots

Trying to factor by splitting the middle term

 3.5     Factoring  x⁸ - x⁴ + 1 

The first term is,  x⁸  its coefficient is  1 .
The middle term is,  -x⁴  its coefficient is  -1 .
The last term, "the constant", is  +1 

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

Step-2 : Find two factors of  1  whose sum equals the coefficient of the middle term, which is   -1 .

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

Trying to factor as a Difference of Squares :

 3.6      Factoring:  x¹²-1 

Check : 1 is the square of 1
Check :  x¹²  is the square of  x⁶ 

Factorization is :       (x⁶ + 1)  •  (x⁶ - 1) 

Trying to factor as a Sum of Cubes :

 3.7      Factoring:  x⁶ + 1 

Check :  1  is the cube of   1 
Check :  x⁶ is the cube of   x²

Factorization is :
             (x² + 1)  •  (x⁴ - x² + 1) 

Polynomial Roots Calculator :

 3.8    Find roots (zeroes) of :       F(x) = x² + 1

     See theory in step 3.4
In this case, the Leading Coefficient is  1  and the Trailing Constant is  1.

 The factor(s) are:

of the Leading Coefficient :  1
 of the Trailing Constant :  1

 Let us test ....

Polynomial Roots Calculator found no rational roots

Trying to factor by splitting the middle term

 3.9     Factoring  x⁴ - x² + 1 

The first term is,  x⁴  its coefficient is  1 .
The middle term is,  -x²  its coefficient is  -1 .
The last term, "the constant", is  +1 

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

Step-2 : Find two factors of  1  whose sum equals the coefficient of the middle term, which is   -1 .

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

Trying to factor as a Difference of Squares :

 3.10      Factoring:  x⁶-1 

Check : 1 is the square of 1
Check :  x⁶  is the square of  x³ 

Factorization is :       (x³ + 1)  •  (x³ - 1) 

Trying to factor as a Sum of Cubes :

 3.11      Factoring:  x³ + 1 

Check :  1  is the cube of   1 
Check :  x³ is the cube of   x¹

Factorization is :
             (x + 1)  •  (x² - x + 1) 

Trying to factor by splitting the middle term

 3.12     Factoring  x² - x + 1 

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

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

Step-2 : Find two factors of  1  whose sum equals the coefficient of the middle term, which is   -1 .

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

Trying to factor as a Difference of Cubes:

 3.13      Factoring:  x³-1 

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 :  1  is the cube of   1 
Check :  x³ is the cube of   x¹

Factorization is :
             (x - 1)  •  (x² + x + 1) 

Trying to factor by splitting the middle term

 3.14     Factoring  x² + x + 1 

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

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

Step-2 : Find two factors of  1  whose sum equals the coefficient of the middle term, which is   1 .

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

Final result :

  2•(x4+1)•(x8-x4+1)•(x2+1)•(x4-x2+1)•(x+1)•(x2-x+1)•(x-1)•(x2+x+1)