Solution - Nonlinear equations

Reformatting the input :

Changes made to your input should not affect the solution:

 (1): "x4"   was replaced by   "x^4". 

Rearrange:

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

                     4^2*x^4-(16)=0 

Step by step solution :

Step  1  :

 1.1     4 = 22

(4)² = (2²)² = 2⁴

Equation at the end of step  1  :

  (24 • x4) -  16  = 0 

Step  2  :

Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   16x⁴ - 16  =   16 • (x⁴ - 1) 

Trying to factor as a Difference of Squares :

 3.2      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 :

 3.3    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 ....

Polynomial Roots Calculator found no rational roots

Trying to factor as a Difference of Squares :

 3.4      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  3  :

  16 • (x2 + 1) • (x + 1) • (x - 1)  = 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.

Equations which are never true :

 4.2      Solve :    16   =  0

This equation has no solution.
A a non-zero constant never equals zero.

Solving a Single Variable Equation :

 4.3      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 i 
                      x=  0.0000 - 1.0000 i 

Solving a Single Variable Equation :

 4.4      Solve  :    x+1 = 0 

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

Solving a Single Variable Equation :

 4.5      Solve  :    x-1 = 0 

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

Four solutions were found :

1.  x = 1
2.  x = -1
3.   x=  0.0000 - 1.0000 i 
   
4.   x=  0.0000 + 1.0000 i