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ALGEBRA
Polynomial functions' coefficients and roots relations, Vietas' formulas
 :: Polynomial functions expressed by zeroes or roots
 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),
thus, the above expression shows following coefficients and roots relations called Vieta's formulas.
Therefore, if the leading coefficient an = 1,
we can write 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.

 :: Graphing polynomial functions given their roots example

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 them 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   where  a = 1 thus, it is the source cubic of the type 2/1 since,  a3a1 > 0.

The 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 )
  so we get,     y - 2 = (x + 1)3 + (x + 1)    or     y = x3 + 3x2 + 4x + 4
as shows the below figure.
 
 
 
 
 
 
 
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