Solution - Factoring binomials using the difference of squares

Reformatting the input :

Changes made to your input should not affect the solution:

 (1): "x7"   was replaced by   "x^7".  1 more similar replacement(s).

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

                     2*x^3-(x^7)=0 

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  2x3 -  x7  = 0 

Step  2  :

Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   2x³ - x⁷  =   -x³ • (x⁴ - 2) 

Trying to factor as a Difference of Squares :

 3.2      Factoring:  x⁴ - 2 

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 : 2 is not a square !!

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

Polynomial Roots Calculator :

 3.3    Find roots (zeroes) of :       F(x) = x⁴ - 2
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  -2.

 The factor(s) are:

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

 Let us test ....

Polynomial Roots Calculator found no rational roots

Equation at the end of step  3  :

  -x3 • (x4 - 2)  = 0 

Step  4  :

Theory - Roots of a product :

 4.1    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 :

 4.2      Solve  :    -x³ = 0 

 Multiply both sides of the equation by (-1) :  x³ = 0


When two things are equal, their cube roots are equal. Taking the cube 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 :

 4.3      Solve  :    x⁴-2 = 0 

 Add  2  to both sides of the equation : 
                      x⁴ = 2
                     x  =  ∜ 2  

 The equation has two real solutions  
 These solutions are  x = ± ∜2 = ± 1.1892  
 

Three solutions were found :

1.  x = ± ∜2 = ± 1.1892
2.  x = 0