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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.    Contents A 