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Let's set up the missing cornerstone!
A revealing insight into the polynomial function
Every polynomial has its initial position at the origin of the coordinate system
There are three methods to transform polynomial expression
f (x) = anxn + an - 1 xn - 1 + an - 2 xn - 2 + . . . + a2x2 + a1x + a0
into its source form
       fs(x) = anxn + an - 2 xn - 2 + an - 3 xn - 3 + . . . + a3x3 + a2x2 + a1x,
whose graph passes through the origin.
Each method is based on the fact that a polynomial written in general form represents translation of its source (original) function in the direction of the coordinate axes, where the coordinates of translations are
Therefore, each polynomial missing second term (an - 1 = 0), represents a source polynomial whose graph is translated in the direction of the y-axis by  y0 = a0.
Let's reveal all three methods.
First method
If we plug the coordinates of translations with changed signs into a given polynomial  y =  f (x), expressed in the general form, i.e.,
  y + y0 = an(x + x0)n + an - 1(x + x0)n - 1 + . . . + a1(x + x0) + a0
then, after expanding and reducing the above expression we get its source function  fs(x).
Inversely, by plugging the coordinates of translations into the source polynomial function, i.e.,
                y - y0 = an(x - x0)n + an - 2(x - x0)n - 2 + . . . + a2(x - x0)2 + a1(x - x0),
after expanding and reducing the above expression, we get given polynomial  f (x).
Note that the complete source polynomial has n - 1 terms, missing second and the absolute term.
Second method
The coefficients of the source polynomial are related to corresponding value of the derivative of the given polynomial at x0, like coefficients of the Taylor polynomial in Taylor's or Maclaurin's formula, thus
where,  an = an,  an - 1 = 0,  a0 =  f (x0), and  f (n - k)(x0) denotes the (n - k)th derivative at x0.
Observe that coefficients of the source polynomial define the value and the direction of the vertical        translation of successive derivatives of given polynomial that is,
  f (n - k)(x0) = (n - k)! an - k .
Further, the horizontal translation of each successive derivative corresponds with x0 of the given                 polynomial.
For example, the coefficient a1 of the source cubic of  f (x) = a3x3 + a2x2 + a1x + a0 
since   f ' (x) = 3a3x2 + 2a2x + a1   and   x0 = - a2/(3a3)  then
Thus,              fs(x) = a3x3 + a1x   is the source cubic polynomial,
where the coefficient a1 equals the slope of the tangent line at the inflection point I (x0, y0).
Further, as    a1 =  f ' (x0),
the coefficient a1 also represents the vertical translation of the first derivative of the cubic polynomial, that is why the sign of the product  a3a1 is used as the additional condition in the classification of the cubic polynomial shown later.
Third method
Sigma notation of the polynomial, where the coefficients of the source polynomial are represented by a recursive formula.
Finally, the polynomial 
 f (x) =  y = anxn + an - 1 xn - 1 + an - 2 xn - 2 + . . . + a2x2 + a1x + a0   we can write
 
                      while,  for   k = 0,             an = an,
     and from an- k,   for   k = n,            a0 = f (x0) = y0.
Thus, the expanded form of the above sum is
y - y0 = an(x - x0)n + an - 2(x - x0)n - 2 + . . . + a2(x - x0)2 + a1(x - x0)
where x0 and y0 are coordinates of translations of the graph of the source polynomial
fs(x) = anxn + an - 2 xn - 2 + . . . + a2x2 + a1x
in the direction of the coordinate axes.
Use of the properties of the polynomial
Therefore, we can write linear function or the polynomial of the first degree
 f (x) = y = a1x + a0   or    y = a1(x - x0)    or    y - y0 = a1x,
where
By setting  x0 = 0  or  y0 = 0  we get  y = a1x  the source linear function.
Polynomial of the second degree or quadratic function
 f (x) = a2x2 + a1x + a0   or     y - y0 = a2(x - x0)2,
where  The turning point T (x0,  y0).
If  x0 = 0 and  y0 = 0  then,  y = a2x2  is the source quadratic. 
Polynomial of the third degree or cubic function
 f (x) = a3x3 + a2x2 + a1x + a0   or     y - y0 = a3(x - x0)3 + a1(x - x0)
where  
By setting  x0 = 0  and  y0 = 0 we get the source cubic function
 y = a3x3 + a1x  where  a1= tan at
therefore, the coefficient  a1 shows the slope of the tangent line at the point of inflection I (x0, y0).
Polynomial of the fourth degree or quartic
 f (x) = a4x4 + a3x3 + a2x2 + a1x + a0   or    y - y0 = a4(x - x0)4 + a2(x - x0)2 + a1(x - x0)
where,  
By setting  x0 = 0  and  y0 = 0 we get
 y = a4x4 + a2x2 + a1x,  the source quartic function.
Thus, we can proceed with, quintic, sextic or an nth degree polynomial.
Benefits the principle provide
Let's show here only a few of the benefits the principle brings.
Classification of the polynomial functions
The first, the principle enables classification of the polynomial functions meaning, it provides basic conditions for particular type of function, or group of functions, an nth degree polynomial includes.
That is, defines the form of the graph specific type of a polynomial function is represented.
Thus for example, cubic polynomial consists of three types, or is represented by three distinct forms (shapes) of graphs of functions.
Using the same approach classified is the quartic polynomial to ten types of characteristic functions and so on.
Each nth degree polynomial is classified using the same basic criteria that include, examination of appearance or exception of each of the source coefficients a, and depending on the sign of their product with the leading coefficient an, since the source coefficients show the direction of the vertical translation of the corresponding derivative.
The classification also defines all variations of a polynomial expression related to the changes of polarity 
of the variables such as,           f (-x),   - f (x)  and   - f (-x) 
that cause the reflections of the graph of the polynomial function around the  y-axis, x-axis and both axes, respectively.
Observe that the first criteria of the classification separates even and odd nth degree polynomials called the power functions or monomials as the first type, since all coefficients a of their source function vanished, (see the above diagram).
Therefore, we write the translated power (or monomial) function or the first type
y - y0 = an(x - x0)n,
where  x0 = - an - 1/(nan),   y0 = f(x0) and n is an even or an odd positive integer.
For  n = 2m, m Î N the even power function has,
  the turning point  T (x0, y0)
  and the real roots if   an y0  < 0,
 
as shows the right figure (where an > 0 ).
For  n = 2m + 1, m Î N the odd power function
  has, the point of inflection I (x0, y0),  at = 0
  and the root
 
as shows the right figure (where an > 0).
The principle provides full control over the polynomial
Hence, by substituting coordinates of translations into the source polynomial function we can move its graph to any desired position on the coordinate plane.
Since an nth degree polynomial with real coefficients can be expressed as a product of its leading coefficient an and n linear factors of the form (x - ri), where ri denotes its real root and/or complex root
f (x) = anxn + an - 1xn - 1 + . . . + a1x + a0 = an(x - r1)(x - r2) . . . (x - rn),
the above expression shows following coefficients and roots relations called Vieta's formulas.
Therefore, if the leading coefficient an = 1, we can write for example,
the quadratic polynomial by roots
      f (x) = x2 + a1x + a0 = (x - r1)(x - r2) = x2 - (r1 + r2)x + r1r2,
the cubic polynomial
      f (x) = x3 + a2x2 + a1x + a0 = (x - r1)(x - r2)(x - r3)    or
      f (x) = x3 - (r1 + r2 + r3)x2 + (r1r2 + r1r3 + r2r3)x - r1r2r3,
the quartic polynomial
      f (x) = x4 + a3x3 + a2x2 + a1x + a0 = (x - r1)(x - r2)(x - r3)(x - r4)    or
      f (x) = x4 - (r1 + r2 + r3 + r4)x3 + (r1r2 + r1r3 + r1r4 + r2r3 + r2r4 + r3r4)x2 -  
                                                                             - (r1r2r3 + r1r2r4 + r1r3r4 + r2r3r4)x + r1r2r3r4,
and so on.
Note that the coefficients of a polynomial are expressed as the alternating sums of corresponding combinations of products of roots.
Example:   Let write the cubic polynomial whose roots are r1 = 1 and  r2,3 = 1 ± i  assuming its leading coefficient a3 = 1, and find its source function and draw their graphs.
Then, find the cubic obtained by moving the source function by  xt = - 1 and  yt = 2, and draw the graph of the translated cubic.
Solution:   We write the cubic polynomial given by its roots to calculate its coefficients, since
                 f (x) = x3 - (r1 + r2 + r3)x2 + (r1r2 + r1r3 + r2r3)x - r1r2r3,
then                  a2 = - (r1 + r2 + r3) = - [1 + (1 + i) + (1 - i)] = - 3,
                        a1 = r1r2 + r1r3 + r2r3 = 1 · (1 + i) + 1 · (1 - i) + (1 + i) · (1 - i) = 4,      
                        a0 = - r1r2r3 = - 1 · (1 + i) · (1 - i) = - 2,
therefore     f (x) = x3 - 3x2 + 4x - 2    or     f (x) = (x - 1)[x - (1 + i)][x - (1 - i)].
To find the source form,   fs(x) = a3x3 + a1x  of the general cubic polynomial we calculate coordinates
of translations x0 and y0,
and plug into      y + y0 = a3(x + x0)3 + a2(x + x0)2 + a1(x + x0) + a0
           thus,              y = (x + 1)3 - 3(x + 1)2 + 4(x + 1) - 2 = x3 + x 
               or         fs(x) = x3 + x   - the source cubic of the type 2/1 since,  a3a1 > 0.
The obtained source cubic we then move to new position by plugging the given coordinates of translations,
       xt = -1 and  yt = 2,      into         y - yt = a3(x - xt )3 + a1(x - xt )
thus,      y - 2 = (x + 1)3 + (x + 1)    or     y = x3 + 3x2 + 4x + 4
as shows the below figure.
Example:   Find the fifth degree polynomial (or quintic) whose roots are,
  r1 = - 1,  r2 = - 0.5,  r3 = 1,  r4 = 1 and  r5 = 3,
assuming its leading coefficient a5 = 1, and find its source function and draw their graphs.
Then, move given function horizontally by xt = - 3 and draw the graph of the translated quintic.
Solution:   Let's calculate coefficients of the quintic using Vieta's formulas,
  a4 = - (r1 + r2 + r3 + r4 + r5) = - 3.5
  a3 = r1r2 + r1r3 + r1r4 + r1r5 + r2r3 + r2r4 + r2r5 + r3r4 + r3r5 + r4r5 = 0
  a2 = - (r1r2r3 + r1r2r4 + r1r2r5 + r1r3r4 + r1r3r5 + r1r4r5 + r2r3r4 + r2r3r5 + r2r4r5 + r3r4r5) = 5
  a1 = r1r2r3r4 + r1r2r3r5 + r1r2r4r5 + r1r3r4r5 + r2r3r4r5 = - 1
  a0 = - r1r2r3r4r5 = - 1.5.
Therefore, the given quintic    y = x5 - 3.5x4 + 5x2 - x - 1.5   is the variant   - f (-x)  of 
f (x) = x5 + 3.5x4 - 5x2 - x + 1.5.
To find the coefficients a of the source quintic polynomial
fs(x) = a5x5 + a3x3 + a2x2 + a1x
we can alternatively use the second method, through direct formula introduced above.
Thus, we should first evaluate the three successive derivatives of the given quintic at x0, divide each value by the corresponding factorial to obtain the three successive coefficients.
Since, then,
Therefore, the source quintic    fs(x) = x5 - 4.9x3 - 1.86x2 + 2.3985x.
To translate given quintic in the direction of the x-axis by xt = - 3, substitute each x in its expression with (x + 3), (what is the same as moving its source function by xt = - 2.3 and  yt = y0 = - 0.42228).
So,  y = x5 - 3.5x4 + 5x2 - x - 1.5  gives  y = (x + 3)5 - 3.5(x + 3)4 + 5(x + 3)2 - (x + 3) - 1.5
    or    y = x5 + 11.5x4 + 48x3 + 86x2 + 56x,  as shows the below figure.
More about polynomial and polynomial functions, an interested visitor can find searching the contents.
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