Solution - Factoring binomials using the difference of squares

Rearrange:

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

                0-(2*x^4+8*x^3-2*x^2-8*x)=0 

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  0-((((2•(x4))+(8•(x3)))-2x2)-8x)  = 0 
 Step  2  :

Equation at the end of step  2  :

  0-((((2•(x4))+23x3)-2x2)-8x)  = 0 
 Step  3  :

Equation at the end of step  3  :

  0 -  (((2x4 +  23x3) -  2x2) -  8x)  = 0 

Step  4  :

Step  5  :

Pulling out like terms :

 5.1     Pull out like factors :

   -2x⁴ - 8x³ + 2x² + 8x  = 

  -2x • (x³ + 4x² - x - 4) 

Checking for a perfect cube :

 5.2    x³ + 4x² - x - 4  is not a perfect cube

Trying to factor by pulling out :

 5.3      Factoring:  x³ + 4x² - x - 4 

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1:  -x - 4 
Group 2:  4x² + x³ 

Pull out from each group separately :

Group 1:   (x + 4) • (-1)
Group 2:   (x + 4) • (x²)
               -------------------
Add up the two groups :
               (x + 4)  •  (x² - 1) 
Which is the desired factorization

Trying to factor as a Difference of Squares :

 5.4      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) 

Equation at the end of step  5  :

  -2x • (x + 1) • (x - 1) • (x + 4)  = 0 

Step  6  :

Theory - Roots of a product :

 6.1    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 :

 6.2      Solve  :    -2x = 0 

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


Divide both sides of the equation by 2:
                     x = 0

Solving a Single Variable Equation :

 6.3      Solve  :    x+1 = 0 

 Subtract  1  from both sides of the equation : 
                      x = -1

Solving a Single Variable Equation :

 6.4      Solve  :    x-1 = 0 

 Add  1  to both sides of the equation : 
                      x = 1

Solving a Single Variable Equation :

 6.5      Solve  :    x+4 = 0 

 Subtract  4  from both sides of the equation : 
                      x = -4

Four solutions were found :

1.  x = -4
2.  x = 1
3.  x = -1
4.  x = 0