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^2)*3*x)^2)-(16*((x^2)*3*x))-(36)=0 

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  (((((x2)•3)•x)2)-(16•(3x2•x)))-36  = 0 
 Step  2  :

Equation at the end of step  2  :

  (((((x2)•3)•x)2)-(24•3x3))-36  = 0 
 Step  3  :

Equation at the end of step  3  :

  (((3x2 • x)2) -  (24•3x3)) -  36  = 0 

Step  4  :

Step  5  :

Pulling out like terms :

 5.1     Pull out like factors :

   9x⁶ - 48x³ - 36  =   3 • (3x⁶ - 16x³ - 12) 

Trying to factor by splitting the middle term

 5.2     Factoring  3x⁶ - 16x³ - 12 

The first term is,  3x⁶  its coefficient is  3 .
The middle term is,  -16x³  its coefficient is  -16 .
The last term, "the constant", is  -12 

Step-1 : Multiply the coefficient of the first term by the constant   3 • -12 = -36 

Step-2 : Find two factors of  -36  whose sum equals the coefficient of the middle term, which is   -16 .

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -18  and  2 
                     3x⁶ - 18x³ + 2x³ - 12

Step-4 : Add up the first 2 terms, pulling out like factors :
                    3x³ • (x³-6)
              Add up the last 2 terms, pulling out common factors :
                    2 • (x³-6)
Step-5 : Add up the four terms of step 4 :
                    (3x³+2)  •  (x³-6)
             Which is the desired factorization

Trying to factor as a Difference of Cubes:

 5.3      Factoring:  x³-6 

Theory : A difference of two perfect cubes,  a³ - b³ can be factored into
              (a-b) • (a² +ab +b²)

Proof :  (a-b)•(a²+ab+b²) =
            a³+a²b+ab²-ba²-b²a-b³ =
            a³+(a²b-ba²)+(ab²-b²a)-b³ =
            a³+0+0-b³ =
            a³-b³

Check :  6  is not a cube !!
Ruling : Binomial can not be factored as the difference of two perfect cubes

Polynomial Roots Calculator :

 5.4    Find roots (zeroes) of :       F(x) = x³-6
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  -6.

 The factor(s) are:

of the Leading Coefficient :  1
 of the Trailing Constant :  1 ,2 ,3 ,6

 Let us test ....

Polynomial Roots Calculator found no rational roots

Trying to factor as a Sum of Cubes :

 5.5      Factoring:  3x³+2 

Theory : A sum of two perfect cubes,  a³ + b³ can be factored into  :
             (a+b) • (a²-ab+b²)
Proof  : (a+b) • (a²-ab+b²) =
    a³-a²b+ab²+ba²-b²a+b³ =
    a³+(a²b-ba²)+(ab²-b²a)+b³=
    a³+0+0+b³=
    a³+b³

Check :  3  is not a cube !!

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

Polynomial Roots Calculator :

 5.6    Find roots (zeroes) of :       F(x) = 3x³+2

     See theory in step 5.4
In this case, the Leading Coefficient is  3  and the Trailing Constant is  2.

 The factor(s) are:

of the Leading Coefficient :  1,3
 of the Trailing Constant :  1 ,2

 Let us test ....

Polynomial Roots Calculator found no rational roots

Equation at the end of step  5  :

  3 • (x3 - 6) • (3x3 + 2)  = 0 

Step  6  :

Theory - Roots of a product :

 6.1    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 :

 6.2      Solve :    3   =  0

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

Solving a Single Variable Equation :

 6.3      Solve  :    x³-6 = 0 

 Add  6  to both sides of the equation : 
                      x³ = 6
When two things are equal, their cube roots are equal. Taking the cube root of the two sides of the equation we get:  
                      x  =  ∛ 6  

 The equation has one real solution
This solution is  x = ∛6 = 1.8171

Solving a Single Variable Equation :

 6.4      Solve  :    3x³+2 = 0 

 Subtract  2  from both sides of the equation : 
                      3x³ = -2
Divide both sides of the equation by 3:
                     x³ = -2/3 = -0.667
When two things are equal, their cube roots are equal. Taking the cube root of the two sides of the equation we get:  
                      x  =  ∛ -2/3  

 Negative numbers have real cube roots.
 ∛ -2/3 = ∛ -1• 2/3  = ∛ -1 • ∛ 2/3  =(-1)•∛ 2/3 

The equation has one real solution, a negative number This solution is  x = ∛ -0.667 = -0.87358

Supplement : Solving Quadratic Equation Directly

Solving    3x6-16x3-12  = 0   directly 

Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula

Solving a Single Variable Equation :

Equations which are reducible to quadratic :

 7.1     Solve   3x⁶-16x³-12 = 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 = x³  transforms the equation into :
 3w²-16w-12 = 0

Solving this new equation using the quadratic formula we get two real solutions :
   6.0000  or  -0.6667

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
   3x⁶-16x³-12 = 0
  are either : 
    x = ∛ 6.000 = 1.8171
   or:
  x = ∛-0.667 = -0.8736

Two solutions were found :

1.  x = ∛ -0.667 = -0.87358
2.  x = ∛6 = 1.8171