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Rational Root Theorem. Possible Rational Roots What are the possible rational roots of Factors of the constant term, 6 , are Factors of the lead coefficient, are Possible rational roots are . Possible Rational Roots. What are the possible rational roots of .

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## Rational Root Theorem

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**Rational Root Theorem**Possible Rational Roots What are the possible rational roots of Factors of the constant term, 6, are Factors of the lead coefficient, are Possible rational roots are**Possible Rational Roots**• What are the possible rational roots of • Factors of the constant term, 8, are • Factors of the lead coefficient, 2, are • Possible rational roots are**Always keep in mind the relationship among zeros, roots, and**x-intercepts. The zeros of a function are the roots, or solutions of the equation . The real zeros, or real roots, are the x-intercepts of the graph of .**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**Find the rational zeros for Find 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 try1. Use synthetic division**1 1 1 –10 8 1 2 –8 1 2 –8 0 1is a zero of the function The depressed polynomial is x2 + 2x – 8 Find the zeros ofx2 + 2x – 8by 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 real zeros of**Find all possible rational zeros of: The possible rational zeros are 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: is a zero of multiplicity two; 2 is a zero; and the other two zeros can be found by solving:**Using The Linear Factorization Theorem**Find a 4th degree polynomial function with real coefficients that has as zeros and such that . Solution: Because is a zero , the conjugate, , must also be a zero. We can now use the Linear Factorization Theorem for a fourth-degree polynomial.**Using The Linear Factorization Theorem**Substituting for in the formula for , we obtain**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**How do we determine the possible number of negative answers? We substitute for every x-value in the equation. Then we look for the sign changes.**1**3 2 Descarte’s Rule of Signs Example Determine the possible number of positive real zeros and negative real zeros of P(x) = x4 – 6x3 + 8x2 + 2x – 1. We first consider the possible number of positive zeros by observing that P(x) has three variations in signs. +x4 – 6x3 + 8x2 + 2x – 1 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). P(x) = (x)4 – 6(x)3 + 8(x)2 + 2(x) 1 = x4 + 6x3 + 8x2 – 2x – 1 Since there is only one variation in sign, P(x) has only one negative real root. Total number of zeros 4 Positive: 3 1 Negative: 1 1 Nonreal: 0 2

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