Step  1  :

Equation at the end of step  1  :

  (2x23 • x) -  2  = 0 
 Step  2  :
 Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   2x²⁴ - 2  =   2 • (x²⁴ - 1) 

Trying to factor as a Difference of Squares :

 3.2      Factoring:  x²⁴ - 1 

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 : 1 is the square of 1
Check :  x²⁴  is the square of  x¹² 

Factorization is :       (x¹² + 1)  •  (x¹² - 1) 

Trying to factor as a Sum of Cubes :

 3.3      Factoring:  x¹² + 1 

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 :  1  is the cube of   1 
Check :  x¹² is the cube of   x⁴

Factorization is :
             (x⁴ + 1)  •  (x⁸ - x⁴ + 1) 

Polynomial Roots Calculator :

 3.4    Find roots (zeroes) of :       F(x) = x⁴ + 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 ....

Polynomial Roots Calculator found no rational roots

Trying to factor by splitting the middle term

 3.5     Factoring  x⁸ - x⁴ + 1 

The first term is,  x⁸  its coefficient is  1 .
The middle term is,  -x⁴  its coefficient is  -1 .
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   -1 .

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

Trying to factor as a Difference of Squares :

 3.6      Factoring:  x¹²-1 

Check : 1 is the square of 1
Check :  x¹²  is the square of  x⁶ 

Factorization is :       (x⁶ + 1)  •  (x⁶ - 1) 

Trying to factor as a Sum of Cubes :

 3.7      Factoring:  x⁶ + 1 

Check :  1  is the cube of   1 
Check :  x⁶ is the cube of   x²

Factorization is :
             (x² + 1)  •  (x⁴ - x² + 1) 

Polynomial Roots Calculator :

 3.8    Find roots (zeroes) of :       F(x) = x² + 1

     See theory in step 3.4
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 ....

Polynomial Roots Calculator found no rational roots

Trying to factor by splitting the middle term

 3.9     Factoring  x⁴ - x² + 1 

The first term is,  x⁴  its coefficient is  1 .
The middle term is,  -x²  its coefficient is  -1 .
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   -1 .

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

Trying to factor as a Difference of Squares :

 3.10      Factoring:  x⁶-1 

Check : 1 is the square of 1
Check :  x⁶  is the square of  x³ 

Factorization is :       (x³ + 1)  •  (x³ - 1) 

Trying to factor as a Sum of Cubes :

 3.11      Factoring:  x³ + 1 

Check :  1  is the cube of   1 
Check :  x³ is the cube of   x¹

Factorization is :
             (x + 1)  •  (x² - x + 1) 

Trying to factor by splitting the middle term

 3.12     Factoring  x² - x + 1 

The first term is,  x²  its coefficient is  1 .
The middle term is,  -x  its coefficient is  -1 .
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   -1 .

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

Trying to factor as a Difference of Cubes:

 3.13      Factoring:  x³-1 

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 :  1  is the cube of   1 
Check :  x³ is the cube of   x¹

Factorization is :
             (x - 1)  •  (x² + x + 1) 

Trying to factor by splitting the middle term

 3.14     Factoring  x² + x + 1 

The first term is,  x²  its coefficient is  1 .
The middle term is,  +x  its coefficient is  1 .
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   1 .

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

Equation at the end of step  3  :

  2•(x4+1)•(x8-x4+1)•(x2+1)•(x4-x2+1)•(x+1)•(x2-x+1)•(x-1)•(x2+x+1)  = 0 

Step  4  :

Theory - Roots of a product :

 4.1    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 :

 4.2      Solve :    2   =  0

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

Solving a Single Variable Equation :

 4.3      Solve  :    x⁴+1 = 0 

 Subtract  1  from both sides of the equation : 
                      x⁴ = -1
                     x  =  ∜ -1  

 The equation has no real solutions. It has 4 imaginary, or complex solutions.

                      x=  0.7071 + 0.7071 i 
                      x=  -0.7071 + 0.7071 i 
                      x=  -0.7071 - 0.7071 i 
                      x=  0.7071 - 0.7071 i 

Solving a Single Variable Equation :

Equations which are reducible to quadratic :

 4.4     Solve   x⁸-x⁴+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 = x⁴  transforms the equation into :
 w²-w+1 = 0

Solving this new equation using the quadratic formula we get two imaginary solutions :
   w =  0.5000 ± 0.8660 i 
Now that we know the value(s) of  w , we can calculate  x  since  x  is  ∜ w  

Since we are speaking 4th root, each of the two imaginary solutions of has 4 roots

Tiger finds these roots using de Moivre's Formula

The 4th roots of   0.500 + 0.866 i   are:

  x =  0.966 + 0.259 i 
  x = -0.259 + 0.966 i 
  x = -0.966 -0.259 i 
  x =  0.259 -0.966 i 

4th roots of   0.500- 0.866 i  :
  x =  0.259  + 0.966 i  
  x = -0.966  + 0.259 i  
  x = -0.259 - 0.966 i 
  x =  0.966 - 0.259 i 
 

Solving a Single Variable Equation :

 4.5      Solve  :    x²+1 = 0 

 Subtract  1  from both sides of the equation : 
                      x² = -1
 
 When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:  
                      x  =  ± √ -1  

 In Math,  i  is called the imaginary unit. It satisfies   i²  =-1. Both   i   and   -i   are the square roots of   -1 
The equation has no real solutions. It has 2 imaginary, or complex solutions.

                      x=  0.0000 + 1.0000 i 
                      x=  0.0000 - 1.0000 i 

Solving a Single Variable Equation :

Equations which are reducible to quadratic :

 4.6     Solve   x⁴-x²+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-w+1 = 0

Solving this new equation using the quadratic formula we get two imaginary solutions :
   w =  0.5000 ± 0.8660 i 
Now that we know the value(s) of  w , we can calculate  x  since  x  is  √ w  

Since we are speaking 2nd root, each of the two imaginary solutions of has 2 roots

Tiger finds these roots using de Moivre's Formula

The 2nd roots of   0.500 + 0.866 i   are:

  x =  0.866 + 0.500 i 
  x = -0.866 -0.500 i 

2nd roots of   0.500- 0.866 i  :
  x = -0.866  + 0.500 i  
  x =  0.866 - 0.500 i 
 

Solving a Single Variable Equation :

 4.7      Solve  :    x+1 = 0 

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

Parabola, Finding the Vertex :

 4.8      Find the Vertex of   y = x²-x+1

Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) .   We know this even before plotting  "y"  because the coefficient of the first term, 1 , is positive (greater than zero). 

 Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two  x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions. 

 Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex. 

 For any parabola,Ax²+Bx+C,the  x -coordinate of the vertex is given by  -B/(2A) . In our case the  x  coordinate is   0.5000  

 Plugging into the parabola formula   0.5000  for  x  we can calculate the  y -coordinate : 
  y = 1.0 * 0.50 * 0.50 - 1.0 * 0.50 + 1.0
or   y = 0.750

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = x²-x+1
Axis of Symmetry (dashed)  {x}={ 0.50} 
Vertex at  {x,y} = { 0.50, 0.75} 
Function has no real roots

Solve Quadratic Equation by Completing The Square

 4.9     Solving   x²-x+1 = 0 by Completing The Square .

 Subtract  1  from both side of the equation :
   x²-x = -1

Now the clever bit: Take the coefficient of  x , which is  1 , divide by two, giving  1/2 , and finally square it giving  1/4 

Add  1/4  to both sides of the equation :
  On the right hand side we have :
   -1  +  1/4    or,  (-1/1)+(1/4) 
  The common denominator of the two fractions is  4   Adding  (-4/4)+(1/4)  gives  -3/4 
  So adding to both sides we finally get :
   x²-x+(1/4) = -3/4

Adding  1/4  has completed the left hand side into a perfect square :
   x²-x+(1/4)  =
   (x-(1/2)) • (x-(1/2))  =
  (x-(1/2))²
Things which are equal to the same thing are also equal to one another. Since
   x²-x+(1/4) = -3/4 and
   x²-x+(1/4) = (x-(1/2))²
then, according to the law of transitivity,
   (x-(1/2))² = -3/4

We'll refer to this Equation as  Eq. #4.9.1  

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of
   (x-(1/2))²   is
   (x-(1/2))^2/2 =
  (x-(1/2))¹ =
   x-(1/2)

Now, applying the Square Root Principle to  Eq. #4.9.1  we get:
   x-(1/2) = √ -3/4

Add  1/2  to both sides to obtain:
   x = 1/2 + √ -3/4
In the set of real numbers, negative numbers do not have square roots. A new set of numbers, called complex, was invented so that negative numbers would have a square root. These numbers are written  (a+b*i) 

Both   i   and   -i   are the square roots of minus 1



Since a square root has two values, one positive and the other negative
   x² - x + 1 = 0
   has two solutions:
  x = 1/2 + √ 3/4 •  i 
   or
  x = 1/2 - √ 3/4 •  i 

Note that  √ 3/4 can be written as
  √ 3  / √ 4   which is √ 3  / 2

Solve Quadratic Equation using the Quadratic Formula

 4.10     Solving    x²-x+1 = 0 by the Quadratic Formula .

 According to the Quadratic Formula,  x  , the solution for   Ax²+Bx+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :
                                     
            - B  ±  √ B²-4AC
  x =   ————————
                      2A

  In our case,  A   =     1
                      B   =    -1
                      C   =   1

Accordingly,  B²  -  4AC   =
                     1 - 4 =
                     -3

Applying the quadratic formula :

               1 ± √ -3
   x  =    —————
                    2

Accordingly,√ -3  = 
                    √ 3 • (-1)  =
                    √ 3  • √ -1   =
                    ±  √ 3  • i

  √ 3   , rounded to 4 decimal digits, is   1.7321
 So now we are looking at:
           x  =  ( 1 ±  1.732 i ) / 2

Two imaginary solutions :

 x =(1+√-3)/2=(1+i√ 3 )/2= 0.5000+0.8660i
  or: 
 x =(1-√-3)/2=(1-i√ 3 )/2= 0.5000-0.8660i

Solving a Single Variable Equation :

 4.11      Solve  :    x-1 = 0 

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

Parabola, Finding the Vertex :

 4.12      Find the Vertex of   y = x²+x+1

Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) .   We know this even before plotting  "y"  because the coefficient of the first term, 1 , is positive (greater than zero). 

 Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two  x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions. 

 Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex. 

 For any parabola,Ax²+Bx+C,the  x -coordinate of the vertex is given by  -B/(2A) . In our case the  x  coordinate is  -0.5000  

 Plugging into the parabola formula  -0.5000  for  x  we can calculate the  y -coordinate : 
  y = 1.0 * -0.50 * -0.50 + 1.0 * -0.50 + 1.0
or   y = 0.750

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = x²+x+1
Axis of Symmetry (dashed)  {x}={-0.50} 
Vertex at  {x,y} = {-0.50, 0.75} 
Function has no real roots