Solution - Equations reducible to quadratic form

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((x6) -  3x3) -  28  = 0 

Step  2  :

Trying to factor by splitting the middle term

 2.1     Factoring  x⁶-3x³-28 

The first term is,  x⁶  its coefficient is  1 .
The middle term is,  -3x³  its coefficient is  -3 .
The last term, "the constant", is  -28 

Step-1 : Multiply the coefficient of the first term by the constant   1 • -28 = -28 

Step-2 : Find two factors of  -28  whose sum equals the coefficient of the middle term, which is   -3 .

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -7  and  4 
                     x⁶ - 7x³ + 4x³ - 28

Step-4 : Add up the first 2 terms, pulling out like factors :
                    x³ • (x³-7)
              Add up the last 2 terms, pulling out common factors :
                    4 • (x³-7)
Step-5 : Add up the four terms of step 4 :
                    (x³+4)  •  (x³-7)
             Which is the desired factorization

Trying to factor as a Sum of Cubes :

 2.2      Factoring:  x³+4 

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

Polynomial Roots Calculator :

 2.3    Find roots (zeroes) of :       F(x) = x³+4
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  4.

 The factor(s) are:

of the Leading Coefficient :  1
 of the Trailing Constant :  1 ,2 ,4

 Let us test ....

Polynomial Roots Calculator found no rational roots

Trying to factor as a Difference of Cubes:

 2.4      Factoring:  x³-7 

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

Polynomial Roots Calculator :

 2.5    Find roots (zeroes) of :       F(x) = x³-7

     See theory in step 2.3
In this case, the Leading Coefficient is  1  and the Trailing Constant is  -7.

 The factor(s) are:

of the Leading Coefficient :  1
 of the Trailing Constant :  1 ,7

 Let us test ....

Polynomial Roots Calculator found no rational roots

Equation at the end of step  2  :

  (x3 + 4) • (x3 - 7)  = 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  :    x³+4 = 0 

 Subtract  4  from both sides of the equation : 
                      x³ = -4
When two things are equal, their cube roots are equal. Taking the cube root of the two sides of the equation we get:  
                      x  =  ∛ -4  

 Negative numbers have real cube roots.
 ∛ -4 = ∛ -1• 4  = ∛ -1 • ∛ 4  =(-1)•∛ 4 

The equation has one real solution, a negative number This solution is  x = -∛4 = -1.5874

Solving a Single Variable Equation :

 3.3      Solve  :    x³-7 = 0 

 Add  7  to both sides of the equation : 
                      x³ = 7
When two things are equal, their cube roots are equal. Taking the cube root of the two sides of the equation we get:  
                      x  =  ∛ 7  

 The equation has one real solution
This solution is  x = ∛7 = 1.9129

Supplement : Solving Quadratic Equation Directly

Solving    x6-3x3-28  = 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 :

 4.1     Solve   x⁶-3x³-28 = 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 :
 w²-3w-28 = 0

Solving this new equation using the quadratic formula we get two real solutions :
   7.0000  or  -4.0000

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
   x⁶-3x³-28 = 0
  are either : 
    x = ∛ 7.000 = 1.9129
   or:
  x = ∛-4.000 = -1.5874

Two solutions were found :

1.  x = ∛7 = 1.9129
2.  x = -∛4 = -1.5874