Rational Root Theorem:



Rational Root Theorem:

If [pic] is in simplest form and is a rational root of the polynomial equation: [pic]with integer coefficients, then p must be a factor of f and q must be a factor of a

Example: (2x – 1)(x + 3) = 2x2 + 5x – 3. The roots (x intercepts) are ½ and -3. When the factors are expanded the 2x and x multiply to = a and are factors of a; and -1 and 3 are factors of f. When you use the factors to get the roots you have something in the form of a factor of f divided by a factor of a. The [pic] has p being factors of f and q as factors of a

Refer to the equation[pic]to answer parts a – c

a) What are all of the possible rational roots

b) Which of those possible roots in part a are roots

c) Use either synthetic of long division to find the remaining roots

Refer to the equation [pic]to answer parts a – c

d) What are all of the possible rational roots

e) Which of those possible roots in part a are roots

f) Use either synthetic of long division to find the remaining roots

Irrational Root Theorem:

If [pic] is a root of a polynomial equation with rational coefficients, then the conjugate [pic] also is a root.

Imaginary Root Theorem:

If [pic] is a root of a polynomial equation with rational coefficients, then the conjugate [pic] also is a root.

Descartes Rule of Signs:

The number of positive real roots of a polynomial equation P(x) = 0, with real coefficients is, is equal to the number of sign changes (either positive to negative or vice versa) between the coefficients of the terms of P(x), when x is positive, or is less than this number by a multiple of two

The number of negative real roots of such a polynomial equation is equal to the number of sign changes (either positive to negative or vice versa) between the coefficients of the terms of P(-x), or is less than this number by a multiple of two

Example: If we have P(x) = (x +5)(x + 1)(x – 2)(x + 2). The roots are at, -5, -1, -2 and 2. There are three real negative roots and 1 real positive root. When we expand the P(x) we get x4 + 6x3 + 5x2 – 24x -20

To apply the Rule of Signs we plug in a positive # for x, count the sign changes between each term and this gives us the number of positive real roots. Than we plug in a negative number for x and count the number of sign changes between terms to get the number of negative roots. I like to use P(1) and P(-1)

P(x) = x4 + 6x3 + 5x2 – 24x - 20

P(1) = + + + - - There is one sign change, there is 1 positive real root

P(-1) = + - + + - There are three sign changes, there 3 negative real roots

Fundamental Theorem of Algebra:

If P(x) is a polynomial of degree [pic] with complex coefficients, then P(x) = 0 as at least one complex root

Corollary:

Including complex roots and multiple roots, an nth degree polynomial has exactly n foots: the related polynomial function has exactly n zeros.

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