polynomials

Division Algorithm
If F(x) and G(x) are polynomials, there exist unique polynomials Q(x) and R(x) such that

    \[ F(x)=Q(x)G(x)+R(x), \]

whether (i) R(x) is the zero polynomial or (ii) deg R < deg G
Remainder Theorem
When a polynomial F(x) is divided by x-a, the remainder is F(a).
Factor Theorem
x-a is a root of the equation F(x)=0 if and only if x-a is a factor of F(x).

The Remainder and Factor Theorems are easy consequences of the Division Algorithm. By setting G(x) = x - a and applying the Division Algorithm, we see that R(x) is constant regardless of whether (i) or (ii) holds. Substituting x = a into F(x) = Q(x)(x - a) + R(x), we see that the constant value of R is F(a). It follows that if F(a) = 0 then x - a is a factor of F(x). Conversely, if x - a is a factor of F(x) then F(a) = 0.

Problem. Find the unique polynomial P of degree three that satisfies P(0) = 0 and P(1) = P(2) = P(3) = 1.
Solution. According to the Factor Theorem, P(x)-1 has factors
x-1, x-2, and x-3. It follows that there is a constant C such that

    \[P(x) = 1 + C(x - 1)(x- 2)(x - 3).\]

To evaluate C, we use the fact that P(0) = 0. Thus

    \[ P(0) = 1 + C(-1)(-2)(-3) = 0, \]

from which we find C = 1/6. Thus the desired polynomial is

    \[P(x)= 1 + \fr{6}(x- 1)(x -2)(x-3).\]


If

    \[(x-r_1)(x-r_2)(x-r_3)\dots(x-r_n)=x^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0\]

then

    \[\fr{r_1}+\fr{r_2}+\dots+\fr{r_n}=-\frac{a_1}{a_0}\]

and the sum of the roots is -a_{n-1}.

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