Rational Root Theorem Finding Zeros of a Polynomial

Rational Root Theorem

Finding Zeros of a Polynomial Function �Use the Rational Zero Theorem to find all possible rational zeros. �Use Synthetic Division to try to find one rational zero (the remainder will be zero). �If “n” is a rational zero, factor the original polynomial as (x – n)q(x). �Test remaining possible rational zeros in q(x). If one is found, then factor again as in the previous step. �Continue in this way until all rational zeros have been found. �See if additional irrational or non-real complex zeros can be found by solving a quadratic equation.

Finding Rational Zeros Possible zeros are + 1, + 2, + 4, + 8 So which one do you pick? Pick any. Find one that is a zero using synthetic division. . .

�Let’s try 1. Use synthetic division 1 1 1 – 10 8 1 2 – 8 0 1 is a zero of the function 2 The depressed polynomial is x + 2 x – 8 Find the zeros of x 2 + 2 x – 8 by factoring or (by using the quadratic formula)… (x + 4)(x – 2) = 0 x = – 4, x = 2 The zeros of f(x) are 1, – 4, and 2

Find all possible rational zeros of: Use synthetic division

Example Continued �This new factor has the same possible rational zeros: �Check to see if -1 is also a zero of this: �Conclusion:

Example Continued �This new factor has as possible rational zeros: �Check to see if -1 is also a zero of this: �Conclusion:

Example Continued �Check to see if 1 is a zero: �Conclusion:

Example Continued �Check to see if 2 is a zero: �Conclusion:

Example Continued �Summary of work done:

Using The Linear Factorization Theorem We can now use the Linear Factorization Theorem for a fourthdegree polynomial.

Using The Linear Factorization Theorem

Descartes Rule of Signs is a method for determining the number of sign changes in a polynomial function.

Descarte’s Rule of Signs and Positive Real Zeros Polynomial Function Sign Changes Conclusion 3 2 1 There is one positive real zero.

Descarte’s Rule of Signs Example Determine the possible number of positive real zeros and negative real zeros of P(x) = x 4 – 6 x 3 + 8 x 2 + 2 x – 1. We first consider the possible number of positive zeros by observing that P(x) has three variations in signs. + x 4 – 6 x 3 + 8 x 2 + 2 x – 1 3 1 2 Thus, by Descartes’ rule of signs, f has either 3 or 3 – 2 = 1 positive real zeros. For negative zeros, consider the variations in signs for P( x) = ( x)4 – 6( x)3 + 8( x)2 + 2( x) 1 Total number of zeros 4 = x 4 + 6 x 3 + 8 x 2 – 2 x – 1 Since there is only one variation in sign, P(x) has only one negative real root. Positive: Negative: Nonreal: 3 1 0 1 1 2