Solution - Other Factorizations

Reformatting the input :

Changes made to your input should not affect the solution:

 (1): "x5"   was replaced by   "x^5". 

Step by step solution :

Step  1  :

Step  2  :

Pulling out like terms :

 2.1     Pull out like factors :

   x⁵ - 9x  =   x • (x⁴ - 9) 

Trying to factor as a Difference of Squares :

 2.2      Factoring:  x⁴ - 9 

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 : 9 is the square of 3
Check :  x⁴  is the square of  x² 

Factorization is :       (x² + 3)  •  (x² - 3) 

Polynomial Roots Calculator :

 2.3    Find roots (zeroes) of :       F(x) = x² + 3
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  3.

 The factor(s) are:

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

 Let us test ....

Polynomial Roots Calculator found no rational roots

Trying to factor as a Difference of Squares :

 2.4      Factoring:  x² - 3 

Check : 3 is not a square !!

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

Equation at the end of step  2  :

  x • (x2 + 3) • (x2 - 3)  = 0 

Step  3  :

Theory - Roots of a product :

 3.1    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 :

 3.2      Solve  :    x = 0 

  Solution is  x = 0

Solving a Single Variable Equation :

 3.3      Solve  :    x²+3 = 0 

 Subtract  3  from both sides of the equation : 
                      x² = -3
 
 When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:  
                      x  =  ± √ -3  

 In Math,  i  is called the imaginary unit. It satisfies   i²  =-1. Both   i   and   -i   are the square roots of   -1 

Accordingly,  √ -3  =
                    √ -1• 3   =
                    √ -1 •√  3   =
                    i •  √ 3

The equation has no real solutions. It has 2 imaginary, or complex solutions.

                      x=  0.0000 + 1.7321 i 
                      x=  0.0000 - 1.7321 i 

Solving a Single Variable Equation :

 3.4      Solve  :    x²-3 = 0 

 Add  3  to both sides of the equation : 
                      x² = 3
 
 When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:  
                      x  =  ± √ 3  

 The equation has two real solutions  
 These solutions are  x = ± √3 = ± 1.7321  
 

5 solutions were found :

1.  x = ± √3 = ± 1.7321
2.   x=  0.0000 - 1.7321 i 
   
3.   x=  0.0000 + 1.7321 i 
   
4.  x = 0