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Solution - Factoring binomials using the difference of squares

x=1
x=1
x=1
x=-1
x=0.00001.0000i
x=0.0000-1.0000i
x=0.0000+1.0000i
x=0.0000+1.0000i
x=0
x=0

Step by Step Solution

Reformatting the input :

Changes made to your input should not affect the solution:

 (1): "x8"   was replaced by   "x^8". 

Rearrange:

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

                     6*(2*x^4)-(3*(4*x^8))=0 

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  (6 • (2 • (x4))) -  (3 • 22x8)  = 0 

Step  2  :

Equation at the end of step  2  :

  (6 • (2 • (x4))) -  (3•22x8)  = 0 

Step  3  :

Equation at the end of step  3  :

  (6 • 2x4) -  (3•22x8)  = 0 

Step  4  :

Multiplying exponents :

 4.1    21  multiplied by  21   = 2(1 + 1) = 22

Equation at the end of step  4  :

  (22•3x4) -  (3•22x8)  = 0 

Step  5  :

Step  6  :

Pulling out like terms :

 6.1     Pull out like factors :

   12x4 - 12x8  =   -12x4 • (x4 - 1) 

Trying to factor as a Difference of Squares :

 6.2      Factoring:  x4 - 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 :  x4  is the square of  x2 

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

Polynomial Roots Calculator :

 6.3    Find roots (zeroes) of :       F(x) = x2 + 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 ....

  P  Q  P/Q  F(P/Q)   Divisor
     -1     1      -1.00      2.00   
     1     1      1.00      2.00   


Polynomial Roots Calculator found no rational roots

Trying to factor as a Difference of Squares :

 6.4      Factoring:  x2 - 1 

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  6  :

  -12x4 • (x2 + 1) • (x + 1) • (x - 1)  = 0 

Step  7  :

Theory - Roots of a product :

 7.1    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 :

 7.2      Solve  :    -12x4 = 0 

 
Multiply both sides of the equation by (-1) :  12x4 = 0


Divide both sides of the equation by 12:
                     x4 = 0
                     x  =  ∜ 0  

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

Solving a Single Variable Equation :

 7.3      Solve  :    x2+1 = 0 

 
Subtract  1  from both sides of the equation : 
 
                     x2 = -1
 
 
When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:  
 
                     x  =  ± √ -1  

 
In Math,  i  is called the imaginary unit. It satisfies   i2  =-1. Both   i   and   -i   are the square roots of   -1 
The equation has no real solutions. It has 2 imaginary, or complex solutions.

                      x=  0.0000 + 1.0000
                      x=  0.0000 - 1.0000

Solving a Single Variable Equation :

 7.4      Solve  :    x+1 = 0 

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

Solving a Single Variable Equation :

 7.5      Solve  :    x-1 = 0 

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

5 solutions were found :

  1.  x = 1
  2.  x = -1
  3.   x=  0.0000 - 1.0000
  4.   x=  0.0000 + 1.0000
  5.  x = 0

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