Polynomial and/or Polynomial Functions and Equations
Polynomial functions
Roots or zeros of polynomial function
Vieta's formulas
Polynomial functions
Roots or zeros of polynomial function
The zeros of a polynomial function are the values of x for which the function equals zero.
That is, the solutions of the equation  f (x) = 0, that are called roots of the polynomial, are the zeros of the polynomial function or the x-intercepts of its graph in a coordinate plane.
At these points the graph of the polynomial function cuts or touches the x-axis.
If the graph of a polynomial intersects with the x-axis at (r, 0), or x = r is a root or zero of a polynomial, then
(x - r) is a factor of that polynomial.
Every polynomial of degree n has exactly n real and/or complex zeros.
An nth degree polynomial has at most n real zeros.
Some of the roots may be repeated. The number of times a root is repeated is called multiplicity or order of the root.
The number ri is a root of the polynomial  f (x) if and only if  f (x) is divisible by (x - ri).
Therefore, a polynomial and/or polynomial function 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, finding the roots of a polynomial f(x) is equivalent to finding its linear divisors or is equivalent to polynomial factorization into linear factors.
Vieta's formulas
Suppose,  r1, r2, r3, · · · , rn,  are roots of a polynomial  f (x) of degree n (counting multiplicities) with leading coefficient an = 1, that is,
f (x) = xn + an-1xn-1 + an-2xn-2 + . . . + a2x2 + a1x + a0
then we can write     f (x) = (x - r1)(x - r2)(x - r3) · · · (x - rn).
By expanding the above expression and collecting like terms, and after comparing corresponding coefficients of both expressions, we get
Vieta's formulas that show coefficients and roots relations.
Note that the coefficients of a polynomial are expressed as the sum of corresponding combinations of roots.
Let write coefficients of 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,
and 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.
Example:  Write the cubic polynomial whose roots are r1 = 1 and  r2,3 = 1 ± i, assuming its leading coefficient a3 = 1. Find its source function and draw their graphs.
Solution:   Let write the cubic polynomial by roots
f (x) = x3 - (r1 + r2 + r3)x2 + (r1r2 + r1r3 + r2r3)x - r1r2r3,
and calculate coefficients
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 should 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.
Alternatively, coefficients of the source polynomial can be calculated like coefficients of the Taylor polynomial
since,  an = an an -1 = 0 and a0 = f (x0), and where  f (n - k) (x0) denotes (n - k)-th derivative at x0.
Hence,  f (x) = x3 - 3x2 + 4x - 2,   f '(x) = 3x2 - 6x + 4 and  f '(x0)  or  f '(1) = 3x2 - 6x + 4 = then
Therefore, the given cubic polynomial is translated only in the direction of the x-axis by x0 = 1, as shows the above picture. Note that the point of inflection of a cubic polynomial I (x0, y0), since f '' (x0) = 0.
Pre-calculus contents E