Chapter 5 Polynomials and Polynomial Functions © Tentinger.

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Presentation transcript:

Chapter 5 Polynomials and Polynomial Functions © Tentinger

Essential Understanding and Objectives Essential Understanding: the factors of a n and a 0 in P(x) can help you factor P(x) and solve the equation P(x) = 0 Objectives Students will be able to: Solve equations using the Rational Root Theorem Use the Conjugate Root Theorem

Iowa Core Curriculum Number and Quantity N.CN.7. Solve quadratic equations with real coefficients that have complex solutions N.CN.8. (+) Extend polynomial identities to the complex numbers

Rational Root Theorem There are a limited number of possible roots of P(x) = 0 Integer roots must be factors of a 0 Rational roots must have reduced form p/q where p is an integer factor of a 0 (constant) and q is an integer factor of a n (leading coefficient). 21x x + 10 = 0 Factors of leading coefficient: 1, 3, 7, 21 Factors of Constant Term: 1, 2, 5, 10 Divide everything by leading coefficient

Example What are the rational roots of 2x 3 – x 2 + 2x + 5 = 0 Plus or minus: 1, 5, ½, 5/2 Make a table, see where the y value will be zero What are the rational roots of 3x 3 + 7x 2 + 6x – 8 = 0 What are the rational roots of 2x 3 +x 2 – 7x – 6 = 0 Go through and test for each root, then you can factor to find other roots.

Conjugate Root Theorem If P(x) is a polynomial with rational coefficients, then irrational roots of P(x) = 0 that have the form a + √b is an irrational root with a and b rational, then a - √b is also a root If a + √b is a root then, a - √b is a root If a +bi is a root then, a – bi is a root A cubic polynomial has real coefficients. If 3 – 2i and 5/2 are two roots, what is one additional root? Constructing a polynomial function What quartic polynomial equation has roots 2 – 3i, 8, 2?

Descartes Rule of Signs The number of positive real roots of P(x) = 0 is either equal to the number of sign changes between consecutive coefficients of P(x) or is less that by an even number The number of negative real roots of P(x) = 0 is either equal to the number of sign changes between consecutive coefficient of P(-x) or is less than that by an even number x 3 – x = 0 2x 4 – x 3 + 3x 2 – 1 = 0