Solution - Finding the roots of polynomials

Step  1  :

             3
 Simplify   ——
            x2

Equation at the end of step  1  :

                    3
  (((x4)+(2•(x2)))-——)+3
                   x2
 Step  2  :

Equation at the end of step  2  :

                     3     
  (((x4) +  2x2) -  ——) +  3
                    x2     

Step  3  :

Rewriting the whole as an Equivalent Fraction :

 3.1   Subtracting a fraction from a whole

Rewrite the whole as a fraction using  x²  as the denominator :

                 x4 + 2x2     (x4 + 2x2) • x2
     x4 + 2x2 =  ————————  =  ———————————————
                    1               x2       

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Step  4  :

Pulling out like terms :

 4.1     Pull out like factors :

   x⁴ + 2x²  =   x² • (x² + 2) 

Polynomial Roots Calculator :

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

Adding fractions that have a common denominator :

 4.3       Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

 x2 • (x2+2) • x2 - (3)     x6 + 2x4 - 3
 ——————————————————————  =  ————————————
           x2                    x2     

Equation at the end of step  4  :

  (x6 + 2x4 - 3)    
  —————————————— +  3
        x2          

Step  5  :

Rewriting the whole as an Equivalent Fraction :

 5.1   Adding a whole to a fraction

Rewrite the whole as a fraction using  x²  as the denominator :

         3     3 • x2
    3 =  —  =  ——————
         1       x2  

Polynomial Roots Calculator :

 5.2    Find roots (zeroes) of :       F(x) = x⁶ + 2x⁴ - 3

     See theory in step 4.2
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 ....

The Factor Theorem states that if P/Q is root of a polynomial then this polynomial can be divided by q*x-p Note that q and p originate from P/Q reduced to its lowest terms

In our case this means that
   x⁶ + 2x⁴ - 3 
can be divided by 2 different polynomials,including by  x - 1 

Polynomial Long Division :

 5.3    Polynomial Long Division
Dividing :  x⁶ + 2x⁴ - 3 
                              ("Dividend")
By         :    x - 1    ("Divisor")

Quotient :  x⁵+x⁴+3x³+3x²+3x+3  Remainder:  0 

Polynomial Roots Calculator :

 5.4    Find roots (zeroes) of :       F(x) = x⁵+x⁴+3x³+3x²+3x+3

     See theory in step 4.2
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 ....

The Factor Theorem states that if P/Q is root of a polynomial then this polynomial can be divided by q*x-p Note that q and p originate from P/Q reduced to its lowest terms

In our case this means that
   x⁵+x⁴+3x³+3x²+3x+3 
can be divided with  x+1 

Polynomial Long Division :

 5.5    Polynomial Long Division
Dividing :  x⁵+x⁴+3x³+3x²+3x+3 
                              ("Dividend")
By         :    x+1    ("Divisor")

Quotient :  x⁴+3x²+3  Remainder:  0 

Trying to factor by splitting the middle term

 5.6     Factoring  x⁴+3x²+3 

The first term is,  x⁴  its coefficient is  1 .
The middle term is,  +3x²  its coefficient is  3 .
The last term, "the constant", is  +3 

Step-1 : Multiply the coefficient of the first term by the constant   1 • 3 = 3 

Step-2 : Find two factors of  3  whose sum equals the coefficient of the middle term, which is   3 .

Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored

Adding fractions that have a common denominator :

 5.7       Adding up the two equivalent fractions

 (x4+3x2+3) • (x+1) • (x-1) + 3 • x2     x6 + 2x4 + 3x2 - 3
 ———————————————————————————————————  =  ——————————————————
                 x2                              x2        

Checking for a perfect cube :

 5.8    x⁶ + 2x⁴ + 3x² - 3  is not a perfect cube

Trying to factor by pulling out :

 5.9      Factoring:  x⁶ + 2x⁴ + 3x² - 3 

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

Group 1:  3x² - 3 
Group 2:  x⁶ + 2x⁴ 

Pull out from each group separately :

Group 1:   (x² - 1) • (3)
Group 2:   (x² + 2) • (x⁴)

Bad news !! Factoring by pulling out fails :

The groups have no common factor and can not be added up to form a multiplication.

Polynomial Roots Calculator :

 5.10    Find roots (zeroes) of :       F(x) = x⁶ + 2x⁴ + 3x² - 3

     See theory in step 4.2
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

Final result :

  x6 + 2x4 + 3x2 - 3
  ——————————————————
          x2