Solution - Equations reducible to quadratic form

Rearrange:

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

                x^6-4*x^4+4*x^2-1-(0/t)=0 

Step by step solution :

Step  1  :

            0
 Simplify   —
            t

Equation at the end of step  1  :

  ((((x6)-(4•(x4)))+(4•(x2)))-1)-0  = 0 
 Step  2  :

Equation at the end of step  2  :

  ((((x6)-(4•(x4)))+22x2)-1)-0  = 0 
 Step  3  :

Equation at the end of step  3  :

  ((((x6) -  22x4) +  22x2) -  1) -  0  = 0 

Step  4  :

Checking for a perfect cube :

 4.1    x⁶-4x⁴+4x²-1  is not a perfect cube

Trying to factor by pulling out :

 4.2      Factoring:  x⁶-4x⁴+4x²-1 

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

Group 1:  4x²-1 
Group 2:  -4x⁴+x⁶ 

Pull out from each group separately :

Group 1:   (4x²-1) • (1)
Group 2:   (x²-4) • (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 :

 4.3    Find roots (zeroes) of :       F(x) = x⁶-4x⁴+4x²-1
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 ....

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
   x⁶-4x⁴+4x²-1 
can be divided by 2 different polynomials,including by  x-1 

Polynomial Long Division :

 4.4    Polynomial Long Division
Dividing :  x⁶-4x⁴+4x²-1 
                              ("Dividend")
By         :    x-1    ("Divisor")

Quotient :  x⁵+x⁴-3x³-3x²+x+1  Remainder:  0 

Polynomial Roots Calculator :

 4.5    Find roots (zeroes) of :       F(x) = x⁵+x⁴-3x³-3x²+x+1

     See theory in step 4.3
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 ....

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+x+1 
can be divided with  x+1 

Polynomial Long Division :

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

Quotient :  x⁴-3x²+1  Remainder:  0 

Trying to factor by splitting the middle term

 4.7     Factoring  x⁴-3x²+1 

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

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

Step-2 : Find two factors of  1  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

Equation at the end of step  4  :

  (x4 - 3x2 + 1) • (x + 1) • (x - 1)  = 0 

Step  5  :

Theory - Roots of a product :

 5.1    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 :

Equations which are reducible to quadratic :

 5.2     Solve   x⁴-3x²+1 = 0

This equation is reducible to quadratic. What this means is that using a new variable, we can rewrite this equation as a quadratic equation Using  w , such that  w = x²  transforms the equation into :
 w²-3w+1 = 0

Solving this new equation using the quadratic formula we get two real solutions :
   2.6180  or   0.3820

Now that we know the value(s) of  w , we can calculate  x  since  x  is  √ w  

Doing just this we discover that the solutions of
   x⁴-3x²+1 = 0
  are either : 
   x =√ 2.618 = 1.61803  or :
   x =√ 2.618 = -1.61803  or :
   x =√ 0.382 = 0.61803  or :
   x =√ 0.382 = -0.61803

Solving a Single Variable Equation :

 5.3      Solve  :    x+1 = 0 

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

Solving a Single Variable Equation :

 5.4      Solve  :    x-1 = 0 

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

6 solutions were found :

1.  x = 1
2.  x = -1
3.  x =√ 0.382 = -0.61803
4.  x =√ 0.382 = 0.61803
5.  x =√ 2.618 = -1.61803
6.  x =√ 2.618 = 1.61803