Solution - Factoring binomials using the difference of squares

Step  1  :

Equation at the end of step  1  :

  (2a23 • a) -  2
 Step  2  :
 Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   2a²⁴ - 2  =   2 • (a²⁴ - 1) 

Trying to factor as a Difference of Squares :

 3.2      Factoring:  a²⁴ - 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 :  a²⁴  is the square of  a¹² 

Factorization is :       (a¹² + 1)  •  (a¹² - 1) 

Trying to factor as a Sum of Cubes :

 3.3      Factoring:  a¹² + 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 :  a¹² is the cube of   a⁴

Factorization is :
             (a⁴ + 1)  •  (a⁸ - a⁴ + 1) 

Polynomial Roots Calculator :

 3.4    Find roots (zeroes) of :       F(a) = a⁴ + 1
Polynomial Roots Calculator is a set of methods aimed at finding values of  a  for which   F(a)=0  

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers  a  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  a⁸ - a⁴ + 1 

The first term is,  a⁸  its coefficient is  1 .
The middle term is,  -a⁴  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:  a¹²-1 

Check : 1 is the square of 1
Check :  a¹²  is the square of  a⁶ 

Factorization is :       (a⁶ + 1)  •  (a⁶ - 1) 

Trying to factor as a Sum of Cubes :

 3.7      Factoring:  a⁶ + 1 

Check :  1  is the cube of   1 
Check :  a⁶ is the cube of   a²

Factorization is :
             (a² + 1)  •  (a⁴ - a² + 1) 

Polynomial Roots Calculator :

 3.8    Find roots (zeroes) of :       F(a) = a² + 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  a⁴ - a² + 1 

The first term is,  a⁴  its coefficient is  1 .
The middle term is,  -a²  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:  a⁶-1 

Check : 1 is the square of 1
Check :  a⁶  is the square of  a³ 

Factorization is :       (a³ + 1)  •  (a³ - 1) 

Trying to factor as a Sum of Cubes :

 3.11      Factoring:  a³ + 1 

Check :  1  is the cube of   1 
Check :  a³ is the cube of   a¹

Factorization is :
             (a + 1)  •  (a² - a + 1) 

Trying to factor by splitting the middle term

 3.12     Factoring  a² - a + 1 

The first term is,  a²  its coefficient is  1 .
The middle term is,  -a  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:  a³-1 

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 :  a³ is the cube of   a¹

Factorization is :
             (a - 1)  •  (a² + a + 1) 

Trying to factor by splitting the middle term

 3.14     Factoring  a² + a + 1 

The first term is,  a²  its coefficient is  1 .
The middle term is,  +a  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•(a4+1)•(a8-a4+1)•(a2+1)•(a4-a2+1)•(a+1)•(a2-a+1)•(a-1)•(a2+a+1)