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Solution - Polynomial long division

x=6
x=6
x=(16-sqrt(192))/2=8-4sqrt(3)=1.072
x=(16-sqrt(192))/2=8-4sqrt(3)=1.072
x=(16+sqrt(192))/2=8+4sqrt(3)=14.928
x=(16+sqrt(192))/2=8+4sqrt(3)=14.928

Other Ways to Solve

Polynomial long division

Step by Step Solution

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((((x4)-(28•(x3)))+(22•61x2))-768x)+576  = 0 

Step  2  :

Equation at the end of step  2  :

  ((((x4) -  (22•7x3)) +  (22•61x2)) -  768x) +  576  = 0 

Step  3  :

Polynomial Roots Calculator :

 3.1    Find roots (zeroes) of :       F(x) = x4-28x3+244x2-768x+576
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  576.

 
The factor(s) are:

of the Leading Coefficient :  1
 
of the Trailing Constant :  1 ,2 ,3 ,4 ,6 ,8 ,9 ,12 ,16 ,18 , etc

 
Let us test ....

  P  Q  P/Q  F(P/Q)   Divisor
     -1     1      -1.00      1617.00   
     -2     1      -2.00      3328.00   
     -3     1      -3.00      5913.00   
     -4     1      -4.00      9600.00   
     -6     1      -6.00     21312.00   
     -8     1      -8.00     40768.00   
     -9     1      -9.00     54225.00   
     -12     1     -12.00     114048.00   
     -16     1     -16.00     255552.00   
     -18     1     -18.00     361728.00   
     1     1      1.00      25.00   
     2     1      2.00      -192.00   
     3     1      3.00      -207.00   
     4     1      4.00      -128.00   
     6     1      6.00      0.00    x-6 
     8     1      8.00      -192.00   
     9     1      9.00      -423.00   
     12     1      12.00     -1152.00   
     16     1      16.00      1600.00   
     18     1      18.00      7488.00   


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
   x4-28x3+244x2-768x+576 
can be divided with  x-6 

Polynomial Long Division :

 3.2    Polynomial Long Division
Dividing :  x4-28x3+244x2-768x+576 
                              ("Dividend")
By         :    x-6    ("Divisor")

dividend  x4 - 28x3 + 244x2 - 768x + 576 
- divisor * x3   x4 - 6x3       
remainder  - 22x3 + 244x2 - 768x + 576 
- divisor * -22x2   - 22x3 + 132x2     
remainder      112x2 - 768x + 576 
- divisor * 112x1       112x2 - 672x   
remainder      - 96x + 576 
- divisor * -96x0       - 96x + 576 
remainder         0

Quotient :  x3-22x2+112x-96  Remainder:  0 

Polynomial Roots Calculator :

 3.3    Find roots (zeroes) of :       F(x) = x3-22x2+112x-96

     See theory in step 3.1
In this case, the Leading Coefficient is  1  and the Trailing Constant is  -96.

 
The factor(s) are:

of the Leading Coefficient :  1
 
of the Trailing Constant :  1 ,2 ,3 ,4 ,6 ,8 ,12 ,16 ,24 ,32 , etc

 
Let us test ....

  P  Q  P/Q  F(P/Q)   Divisor
     -1     1      -1.00      -231.00   
     -2     1      -2.00      -416.00   
     -3     1      -3.00      -657.00   
     -4     1      -4.00      -960.00   
     -6     1      -6.00     -1776.00   
     -8     1      -8.00     -2912.00   
     -12     1     -12.00     -6336.00   
     -16     1     -16.00     -11616.00   
     -24     1     -24.00     -29280.00   
     -32     1     -32.00     -58976.00   
     1     1      1.00      -5.00   
     2     1      2.00      48.00   
     3     1      3.00      69.00   
     4     1      4.00      64.00   
     6     1      6.00      0.00    x-6 
     8     1      8.00      -96.00   
     12     1      12.00      -192.00   
     16     1      16.00      160.00   
     24     1      24.00      3744.00   
     32     1      32.00     13728.00   


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
   x3-22x2+112x-96 
can be divided with  x-6 

Polynomial Long Division :

 3.4    Polynomial Long Division
Dividing :  x3-22x2+112x-96 
                              ("Dividend")
By         :    x-6    ("Divisor")

dividend  x3 - 22x2 + 112x - 96 
- divisor * x2   x3 - 6x2     
remainder  - 16x2 + 112x - 96 
- divisor * -16x1   - 16x2 + 96x   
remainder      16x - 96 
- divisor * 16x0       16x - 96 
remainder       0

Quotient :  x2-16x+16  Remainder:  0 

Trying to factor by splitting the middle term

 3.5     Factoring  x2-16x+16 

The first term is,  x2  its coefficient is  1 .
The middle term is,  -16x  its coefficient is  -16 .
The last term, "the constant", is  +16 

Step-1 : Multiply the coefficient of the first term by the constant   1 • 16 = 16 

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

     -16   +   -1   =   -17
     -8   +   -2   =   -10
     -4   +   -4   =   -8
     -2   +   -8   =   -10
     -1   +   -16   =   -17
     1   +   16   =   17
     2   +   8   =   10
     4   +   4   =   8
     8   +   2   =   10
     16   +   1   =   17


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

Multiplying Exponential Expressions :

 3.6    Multiply  (x-6)  by  (x-6) 

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

In our case, the common base is  (x-6)  and the exponents are :
          1 , as  (x-6)  is the same number as  (x-6)1 
 and   1 , as  (x-6)  is the same number as  (x-6)1 
The product is therefore,  (x-6)(1+1) = (x-6)2 

Equation at the end of step  3  :

  (x2 - 16x + 16) • (x - 6)2  = 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.

Parabola, Finding the Vertex :

 4.2      Find the Vertex of   y = x2-16x+16

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   8.0000  

 
Plugging into the parabola formula   8.0000  for  x  we can calculate the  y -coordinate : 
 
 y = 1.0 * 8.00 * 8.00 - 16.0 * 8.00 + 16.0
or   y = -48.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = x2-16x+16
Axis of Symmetry (dashed)  {x}={ 8.00} 
Vertex at  {x,y} = { 8.00,-48.00} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = { 1.07, 0.00} 
Root 2 at  {x,y} = {14.93, 0.00} 

Solve Quadratic Equation by Completing The Square

 4.3     Solving   x2-16x+16 = 0 by Completing The Square .

 
Subtract  16  from both side of the equation :
   x2-16x = -16

Now the clever bit: Take the coefficient of  x , which is  16 , divide by two, giving  8 , and finally square it giving  64 

Add  64  to both sides of the equation :
  On the right hand side we have :
   -16  +  64    or,  (-16/1)+(64/1) 
  The common denominator of the two fractions is  1   Adding  (-16/1)+(64/1)  gives  48/1 
  So adding to both sides we finally get :
   x2-16x+64 = 48

Adding  64  has completed the left hand side into a perfect square :
   x2-16x+64  =
   (x-8) • (x-8)  =
  (x-8)2
Things which are equal to the same thing are also equal to one another. Since
   x2-16x+64 = 48 and
   x2-16x+64 = (x-8)2
then, according to the law of transitivity,
   (x-8)2 = 48

We'll refer to this Equation as  Eq. #4.3.1  

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

Note that the square root of
   (x-8)2   is
   (x-8)2/2 =
  (x-8)1 =
   x-8


Now, applying the Square Root Principle to  Eq. #4.3.1  we get:
   x-8 = 48

Add  8  to both sides to obtain:
   x = 8 + √ 48

Since a square root has two values, one positive and the other negative
   x2 - 16x + 16 = 0
   has two solutions:
  x = 8 + √ 48
   or
  x = 8 - √ 48

Solve Quadratic Equation using the Quadratic Formula

 4.4     Solving    x2-16x+16 = 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   =   -16
                      C   =   16

Accordingly,  B2  -  4AC   =
                     256 - 64 =
                     192

Applying the quadratic formula :

               16 ± √ 192
   x  =    ——————
                      2

Can  √ 192 be simplified ?

Yes!   The prime factorization of  192   is
   2•2•2•2•2•2•3 
To be able to remove something from under the radical, there have to be  2  instances of it (because we are taking a square i.e. second root).

192   =  √ 2•2•2•2•2•2•3   =2•2•2•√ 3   =
                ±  8 • √ 3


  √ 3   , rounded to 4 decimal digits, is   1.7321
 So now we are looking at:
           x  =  ( 16 ± 8 •  1.732 ) / 2

Two real solutions:

 x =(16+√192)/2=8+4√ 3 = 14.928

or:

 x =(16-√192)/2=8-4√ 3 = 1.072

Solving a Single Variable Equation :

 4.5      Solve  :    (x-6)2 = 0 

 
 (x-6) 2 represents, in effect, a product of 2 terms which is equal to zero

For the product to be zero, at least one of these terms must be zero. Since all these terms are equal to each other, it actually means :   x-6  = 0

Add  6  to both sides of the equation : 
 
                     x = 6

Three solutions were found :

  1.  x = 6
  2.  x =(16-√192)/2=8-4√ 3 = 1.072
  3.  x =(16+√192)/2=8+4√ 3 = 14.928

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