Solution - Factoring binomials using the difference of squares

Reformatting the input :

Changes made to your input should not affect the solution:

 (1): "x6"   was replaced by   "x^6". 

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  (2 • (x2)) -  (22•3x6)  = 0 
 Step  2  :

Equation at the end of step  2  :

  2x2 -  (22•3x6)  = 0 

Step  3  :

Step  4  :

Pulling out like terms :

 4.1     Pull out like factors :

   2x² - 12x⁶  =   -2x² • (6x⁴ - 1) 

Trying to factor as a Difference of Squares :

 4.2      Factoring:  6x⁴ - 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 :  6  is not a square !!

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

Polynomial Roots Calculator :

 4.3    Find roots (zeroes) of :       F(x) = 6x⁴ - 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  6  and the Trailing Constant is  -1.

 The factor(s) are:

of the Leading Coefficient :  1,2 ,3 ,6
 of the Trailing Constant :  1

 Let us test ....

Polynomial Roots Calculator found no rational roots

Equation at the end of step  4  :

  -2x2 • (6x4 - 1)  = 0 

Step  5  :

Theory - Roots of a product :

 5.1    A product of several terms equals zero. 

 When a product of two or more terms equals zero, then at least one of the terms must be zero. 

 We shall now solve each term = 0 separately 

 In other words, we are going to solve as many equations as there are terms in the product 

 Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

 5.2      Solve  :    -2x² = 0 

 Multiply both sides of the equation by (-1) :  2x² = 0


Divide both sides of the equation by 2:
                     x² = 0
 
 When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:  
                      x  =  ± √ 0  

 Any root of zero is zero. This equation has one solution which is  x = 0

Solving a Single Variable Equation :

 5.3      Solve  :    6x⁴-1 = 0 

 Add  1  to both sides of the equation : 
                      6x⁴ = 1
Divide both sides of the equation by 6:
                     x⁴ = 1/6 = 0.167
                     x  =  ∜ 1/6  

 The equation has two real solutions  
 These solutions are  x = ∜ 0.167 = ± 0.63894  
 

Three solutions were found :

1.  x = ∜ 0.167 = ± 0.63894
2.  x = 0