Solution - Approximation

Step by step solution :

Step  1  :

              18000 
 Simplify   ————————
            (1 + x)5

Equation at the end of step  1  :

             18000 
  -3000 +  ————————  = 0 
           (x + 1)5

Step  2  :

Rewriting the whole as an Equivalent Fraction :

 2.1   Adding a fraction to a whole

Rewrite the whole as a fraction using  (x+1)⁵   as the denominator :

              -3000     -3000 • (x + 1)5
     -3000 =  —————  =  ————————————————
                1           (x + 1)5    

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 :

 2.2       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:

 -3000 • (x+1)5 + 18000     -3000x5 - 15000x4 - 30000x3 - 30000x2 - 15000x + 15000
 ——————————————————————  =  ——————————————————————————————————————————————————————
       1 • (x+1)5                   1 • (x5 + 5x4 + 10x3 + 10x2 + 5x + 1)         

Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   -3000x⁵ - 15000x⁴ - 30000x³ - 30000x² - 15000x + 15000  = 

  -3000 • (x⁵ + 5x⁴ + 10x³ + 10x² + 5x - 5) 

Trying to factor by pulling out :

 3.2      Factoring:  x⁵ + 5x⁴ + 10x³ + 10x² + 5x - 5 

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

Group 1:  10x³ + 10x² 
Group 2:  5x⁴ + x⁵ 
Group 3:  5x - 5 

Pull out from each group separately :

Group 1:   (x + 1) • (10x²)
Group 2:   (x + 5) • (x⁴)
Group 3:   (x - 1) • (5)


Looking for common sub-expressions :

Group 1:   (x + 1) • (10x²)
Group 3:   (x - 1) • (5)
Group 2:   (x + 5) • (x⁴)

Bad news !! Factoring by pulling out fails :

The groups have no common factor and can not be added up to form a multiplication.

Trying to factor by pulling out :

 3.3      Factoring:  x⁵ + 5x⁴ + 10x³ + 10x² + 5x + 1 

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

Group 1:  10x³ + 10x² 
Group 2:  5x⁴ + x⁵ 
Group 3:  5x + 1 

Pull out from each group separately :

Group 1:   (x + 1) • (10x²)
Group 2:   (x + 5) • (x⁴)
Group 3:   (5x + 1) • (1)


Looking for common sub-expressions :

Group 1:   (x + 1) • (10x²)
Group 3:   (5x + 1) • (1)
Group 2:   (x + 5) • (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 :

 3.4    Find roots (zeroes) of :       F(x) = x⁵ + 5x⁴ + 10x³ + 10x² + 5x - 5
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  -5.

 The factor(s) are:

of the Leading Coefficient :  1
 of the Trailing Constant :  1 ,5

 Let us test ....

Polynomial Roots Calculator found no rational roots

Polynomial Roots Calculator :

 3.5    Find roots (zeroes) of :       F(x) = x⁵ + 5x⁴ + 10x³ + 10x² + 5x + 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 ....

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 + 5x⁴ + 10x³ + 10x² + 5x + 1 
can be divided with  x + 1 

Polynomial Long Division :

 3.6    Polynomial Long Division
Dividing :  x⁵ + 5x⁴ + 10x³ + 10x² + 5x + 1 
                              ("Dividend")
By         :    x + 1    ("Divisor")

Quotient :  x⁴+4x³+6x²+4x+1  Remainder:  0 

Polynomial Roots Calculator :

 3.7    Find roots (zeroes) of :       F(x) = x⁴+4x³+6x²+4x+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 ....

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
   x⁴+4x³+6x²+4x+1 
can be divided with  x+1 

Polynomial Long Division :

 3.8    Polynomial Long Division
Dividing :  x⁴+4x³+6x²+4x+1 
                              ("Dividend")
By         :    x+1    ("Divisor")

Quotient :  x³+3x²+3x+1  Remainder:  0 

Polynomial Roots Calculator :

 3.9    Find roots (zeroes) of :       F(x) = x³+3x²+3x+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 ....

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+3x²+3x+1 
can be divided with  x+1 

Polynomial Long Division :

 3.10    Polynomial Long Division
Dividing :  x³+3x²+3x+1 
                              ("Dividend")
By         :    x+1    ("Divisor")

Quotient :  x²+2x+1  Remainder:  0 

Trying to factor by splitting the middle term

 3.11     Factoring  x²+2x+1 

The first term is,  x²  its coefficient is  1 .
The middle term is,  +2x  its coefficient is  2 .
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   2 .

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  1  and  1 
                     x² + 1x + 1x + 1

Step-4 : Add up the first 2 terms, pulling out like factors :
                    x • (x+1)
              Add up the last 2 terms, pulling out common factors :
                     1 • (x+1)
Step-5 : Add up the four terms of step 4 :
                    (x+1)  •  (x+1)
             Which is the desired factorization

Multiplying Exponential Expressions :

 3.12    Multiply  (x+1)  by  (x+1) 

The rule says : To multiply exponential expressions which have the same base, add up their exponents.

In our case, the common base is  (x+1)  and the exponents are :
          1 , as  (x+1)  is the same number as  (x+1)¹ 
 and   1 , as  (x+1)  is the same number as  (x+1)¹ 
The product is therefore,  (x+1)⁽¹⁺¹⁾ = (x+1)² 

Multiplying Exponential Expressions :

 3.13    Multiply  (x+1)²   by  (x+1) 

The rule says : To multiply exponential expressions which have the same base, add up their exponents.

In our case, the common base is  (x+1)  and the exponents are :
          2
 and   1 , as  (x+1)  is the same number as  (x+1)¹ 
The product is therefore,  (x+1)⁽²⁺¹⁾ = (x+1)³ 

Multiplying Exponential Expressions :

 3.14    Multiply  (x+1)³   by  (x+1) 

The rule says : To multiply exponential expressions which have the same base, add up their exponents.

In our case, the common base is  (x+1)  and the exponents are :
          3
 and   1 , as  (x+1)  is the same number as  (x+1)¹ 
The product is therefore,  (x+1)⁽³⁺¹⁾ = (x+1)⁴ 

Multiplying Exponential Expressions :

 3.15    Multiply  (x+1)⁴   by  (x+1) 

The rule says : To multiply exponential expressions which have the same base, add up their exponents.

In our case, the common base is  (x+1)  and the exponents are :
          4
 and   1 , as  (x+1)  is the same number as  (x+1)¹ 
The product is therefore,  (x+1)⁽⁴⁺¹⁾ = (x+1)⁵ 

Equation at the end of step  3  :

  -3000 • (x5 + 5x4 + 10x3 + 10x2 + 5x - 5) 
  —————————————————————————————————————————  = 0 
                  (x + 1)5                 

Step  4  :

When a fraction equals zero :

 4.1    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:

  -3000•(x5+5x4+10x3+10x2+5x-5) 
  ————————————————————————————— • (x+1)5 = 0 • (x+1)5
             (x+1)5            

Now, on the left hand side, the  (x+1)⁵   cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   -3000  •  (x⁵+5x⁴+10x³+10x²+5x-5)  = 0

Equations which are never true :

 4.2      Solve :    -3000   =  0

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

Equations of order 5 or higher :

 4.3     Solve   x⁵+5x⁴+10x³+10x²+5x-5 = 0

Points regarding equations of degree five or higher.

 (1)  There is no general method (Formula) for solving polynomial equations of degree five or higher.

 (2)  By the Fundamental theorem of Algebra, if we allow complex numbers, an equation of degree  n  will have exactly  n  solutions
(This is if we count double solutions as  2 , triple solutions as  3  and so on

) (3)  By the Abel-Ruffini theorem, the solutions can not always be presented in the conventional way using only a finite amount of additions, subtractions, multiplications, divisions or root extractions

 (4)  If  F(x)  is a polynomial of odd degree with real coefficients, then the equation  F(X)=0  has at least one real solution.

 (5)  Using methods such as the  Bisection  Method, real solutions can be approximated to any desired degree of accuracy.

Approximating a root using the Bisection Method :

We now use the Bisection Method to approximate one of the solutions. The Bisection Method is an iterative procedure to approximate a root (Root is another name for a solution of an equation).

The function is   F(x) = x⁵ + 5x⁴ + 10x³ + 10x² + 5x - 5

At   x=   0.00   F(x)  is equal to  -5.00 
At   x=   1.00   F(x)  is equal to  26.00 

Intuitively we feel, and justly so, that since  F(x)  is negative on one side of the interval, and positive on the other side then, somewhere inside this interval,  F(x)  is zero

Procedure :
(1) Find a point "Left" where F(Left) < 0

(2) Find a point 'Right' where F(Right) > 0

(3) Compute 'Middle' the middle point of the interval [Left,Right]

(4) Calculate Value = F(Middle)

(5) If Value is close enough to zero goto Step (7)

Else :
If Value < 0 then : Left <- Middle
If Value > 0 then : Right <- Middle

(6) Loop back to Step (3)

(7) Done!! The approximation found is Middle

Follow Middle movements to understand how it works :

    Left       Value(Left)     Right       Value(Right)

 0.000000000   -5.000000000  1.000000000   26.000000000
 0.000000000   -5.000000000  1.000000000   26.000000000
 0.000000000   -5.000000000  0.500000000    1.593750000
 0.250000000   -2.948242188  0.500000000    1.593750000
 0.375000000   -1.085113525  0.500000000    1.593750000
 0.375000000   -1.085113525  0.437500000    0.138175011
 0.406250000   -0.500633329  0.437500000    0.138175011
 0.421875000   -0.188247760  0.437500000    0.138175011
 0.429687500   -0.026820026  0.437500000    0.138175011
 0.429687500   -0.026820026  0.433593750    0.055227920
 0.429687500   -0.026820026  0.431640625    0.014092013
 0.430664062   -0.006391933  0.431640625    0.014092013
 0.430664062   -0.006391933  0.431152344    0.003843051
 0.430908203   -0.001276187  0.431152344    0.003843051
 0.430908203   -0.001276187  0.431030273    0.001282996
 0.430908203   -0.001276187  0.430969238    0.000003295
 0.430938721   -0.000636473  0.430969238    0.000003295
 0.430953979   -0.000316596  0.430969238    0.000003295
 0.430961609   -0.000156652  0.430969238    0.000003295
 0.430965424   -0.000076679  0.430969238    0.000003295
 0.430967331   -0.000036692  0.430969238    0.000003295
 0.430968285   -0.000016698  0.430969238    0.000003295
 0.430968761   -0.000006702  0.430969238    0.000003295

     Next Middle will get us close enough to zero:

     F(  0.430969119 ) is   0.000000796  

     The desired approximation of the solution is:

       x ≓ 0.430969119

     Note, ≓ is the approximation symbol

Supplement : Solving Quadratic Equation Directly

Solving    x2 + 2x + 1  = 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

Parabola, Finding the Vertex :

 5.1      Find the Vertex of   y = x²+2x+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  -1.0000  

 Plugging into the parabola formula  -1.0000  for  x  we can calculate the  y -coordinate : 
  y = 1.0 * -1.00 * -1.00 + 2.0 * -1.00 + 1.0
or   y = 0.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = x²+2x+1
Vertex at  {x,y} = {-1.00, 0.00} 
x-Intercept (Root) :
One Root at  {x,y}={-1.00, 0.00} 
Note that the root coincides with
the Vertex and the Axis of Symmetry
coinsides with the line  x = 0 

Solve Quadratic Equation by Completing The Square

 5.2     Solving   x²+2x+1 = 0 by Completing The Square .

 Subtract  1  from both side of the equation :
   x²+2x = -1

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

Add  1  to both sides of the equation :
  On the right hand side we have :
   -1  +  1    or,  (-1/1)+(1/1) 
  The common denominator of the two fractions is  1   Adding  (-1/1)+(1/1)  gives  0/1 
  So adding to both sides we finally get :
   x²+2x+1 = 0

Adding  1  has completed the left hand side into a perfect square :
   x²+2x+1  =
   (x+1) • (x+1)  =
  (x+1)²
Things which are equal to the same thing are also equal to one another. Since
   x²+2x+1 = 0 and
   x²+2x+1 = (x+1)²
then, according to the law of transitivity,
   (x+1)² = 0

We'll refer to this Equation as  Eq. #5.2.1  

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   is
   (x+1)^2/2 =
  (x+1)¹ =
   x+1

Now, applying the Square Root Principle to  Eq. #5.2.1  we get:
   x+1 = √ 0

Subtract  1  from both sides to obtain:
   x = -1 + √ 0
The square root of zero is zero

This quadratic equation has one solution only. That's because adding zero is the same as subtracting zero.

The solution is:
   x  =  -1 

Solve Quadratic Equation using the Quadratic Formula

 5.3     Solving    x²+2x+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   =    2
                      C   =   1

Accordingly,  B²  -  4AC   =
                     4 - 4 =
                     0

Applying the quadratic formula :

               -2 ± √ 0
   x  =    —————
                    2

The square root of zero is zero

This quadratic equation has one solution only. That's because adding zero is the same as subtracting zero.

The solution is:
  x = -2 / 2 = -1

One solution was found :