Solution - Finding the roots of polynomials

Rearrange:

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

                     x-1/x^2-(x/x-3)=0 

Step by step solution :

Step  1  :

            x
 Simplify   —
            x

Equation at the end of step  1  :

          1            
  (x -  ————) -  (1 -  3)  = 0 
        (x2)           
 Step  2  :
             1
 Simplify   ——
            x2

Equation at the end of step  2  :

         1     
  (x -  ——) -  -2  = 0 
        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 :

          x     x • x2
     x =  —  =  ——————
          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

Adding fractions that have a common denominator :

 3.2       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:

 x • x2 - (1)     x3 - 1
 ————————————  =  ——————
      x2            x2  

Equation at the end of step  3  :

  (x3 - 1)    
  ———————— -  -2  = 0 
     x2       

Step  4  :

Rewriting the whole as an Equivalent Fraction :

 4.1   Subtracting a whole from a fraction

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

          -2     -2 • x2
    -2 =  ——  =  ———————
          1        x2   

Trying to factor as a Difference of Cubes:

 4.2      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

 4.3     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

Adding fractions that have a common denominator :

 4.4       Adding up the two equivalent fractions

 (x-1) • (x2+x+1) - (-2 • x2)     x3 + 2x2 - 1
 ————————————————————————————  =  ————————————
              x2                       x2     

Polynomial Roots Calculator :

 4.5    Find roots (zeroes) of :       F(x) = x³ + 2x² - 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
   x³ + 2x² - 1 
can be divided with  x + 1 

Polynomial Long Division :

 4.6    Polynomial Long Division
Dividing :  x³ + 2x² - 1 
                              ("Dividend")
By         :    x + 1    ("Divisor")

Quotient :  x²+x-1  Remainder:  0 

Trying to factor by splitting the middle term

 4.7     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  4  :

  (x2 + x - 1) • (x + 1)
  ——————————————————————  = 0 
            x2          

Step  5  :

When a fraction equals zero :

 5.1    When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

  (x2+x-1)•(x+1)
  —————————————— • x2 = 0 • x2
        x2      

Now, on the left hand side, the  x²  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   (x²+x-1)  •  (x+1)  = 0

Theory - Roots of a product :

 5.2    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.

Parabola, Finding the Vertex :

 5.3      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 = -1.250

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,-1.25} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = {-1.62, 0.00} 
Root 2 at  {x,y} = { 0.62, 0.00} 

Solve Quadratic Equation by Completing The Square

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

 Add  1  to 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  5/4 
  So adding to both sides we finally get :
   x²+x+(1/4) = 5/4

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) = 5/4 and
   x²+x+(1/4) = (x+(1/2))²
then, according to the law of transitivity,
   (x+(1/2))² = 5/4

We'll refer to this Equation as  Eq. #5.4.1  

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. #5.4.1  we get:
   x+(1/2) = √ 5/4

Subtract  1/2  from both sides to obtain:
   x = -1/2 + √ 5/4

Since a square root has two values, one positive and the other negative
   x² + x - 1 = 0
   has two solutions:
  x = -1/2 + √ 5/4
   or
  x = -1/2 - √ 5/4

Note that  √ 5/4 can be written as
  √ 5  / √ 4   which is √ 5  / 2

Solve Quadratic Equation using the Quadratic Formula

 5.5     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) =
                     5

Applying the quadratic formula :

               -1 ± √ 5
   x  =    —————
                    2

  √ 5   , rounded to 4 decimal digits, is   2.2361
 So now we are looking at:
           x  =  ( -1 ±  2.236 ) / 2

Two real solutions:

 x =(-1+√5)/2= 0.618

or:

 x =(-1-√5)/2=-1.618

Solving a Single Variable Equation :

 5.6      Solve  :    x+1 = 0 

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

Three solutions were found :

1.  x = -1
2.  x =(-1-√5)/2=-1.618
3.  x =(-1+√5)/2= 0.618