Describe how both the rational root theorem and descartes’ rules of signs you to find the zeros of a polynomial? give me an example that shows how to apply these concepts. how was the fundamental theorem of algebra used in the process to find the zeros?

Step-by-step explanation:

Rational Root Theorem: If the polynomial

P(x) = a n x n + a n – 1 x n – 1 + ... + a 2 x 2 + a 1 x + a 0

has any rational roots, then they must be of the form of (factors of a0/factors of an).

Example: F(x) = 4x² + 5x +2

If this polynomial has any rational roots, then they must be (factors of 2)/(factors of 4), so (±1, ±2)/(±1, ±2, ±4). So if this polynomial has any rational roots, they must be either: ±1, ±1/2, ±1/4, or ±2. Notice that this polynomial doesn't have to have any rational roots, but if it does, then the roots must fit the Rational Root Theorem.

Descartes' Rules of Signs:

a). In a polynomial, how many time the sign changes is how many positive roots the polynomial will have.

Example: 5x³ + 6x² - 2x - 1

In this expression, the sign only changed once, between 6x² and 2x, so it will only have one positive root.

Example 2: 6x³ - 4x² + x - 6

In this expression, the sign changed 3 times (remember there is a invisible "+" sign before the 6x³), so it will have 3 positive roots.

b). In a polynomial, if you plug in "-x" for all "x", then how many times the new polynomial changes sign is how many negative roots the old polynomial have.

Example: 5x³ + 6x² - 2x - 1.

If we plug in "-x" for all "x", then we get 5(-x)³ + 6(-x)² - 2(-x) -1, which simplifies to -5x³ + 6x² + 2x -1. In this new expression, the sign changed twice, so we have two negative roots for the expression. Notice how we got one positive root the first time and two negative roots the second time, and 1 + 2 = 3. The Fundamental Theorem of Algebra states that for a nth degree polynomial, it will have n complex roots. The polynomial we worked with was a 3rd degree polynomial, and we got 1 + 2 = 3 roots in the end.

1. minimum

2. 6400

3. eight

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