3.6 Polynomial Functions Part 2 Chapter 3 3.6 Polynomial Functions Part 2

Objectives: SWBAT: Find the roots of a polynomial function analytically given at least one of the roots. HW: p. 223: 34-44 (even); 46-54 (even)

Remainders and Factors: If the remainder is 0 when one polynomial is divided by another polynomial, the divisor and the quotient are factors of the polynomial.

Remainder Theorem: The remainder when f(x) is divided by (x – c) is f(c).

Students Try!!!

Factor Theorem:

Fundamental Polynomial Connections: Let f(x) be a polynomial. If r is a real number that satisfies any of the following conditions, then r satisfies all the statements: r is a zero of the function f. r is an x-intercept of the graph of the function f. x = r is a solution, or root, of the equation f(x) = 0. x – r is a factor of the polynomial f(x).

Example: One of the zeros of f(x) = 3x3 – 2x2 – 7x – 2, is 2. Find: all the zeros of f the x-intercepts of the graph of f the solutions to 3x3 – 2x2 – 7x – 2 = 0 the linear factors with real coefficients of 3x3 – 2x2 – 7x – 2

Students Try!!! Answers: the zeros of f 2, –⅓, –1 the x-intercepts of the graph of f (2, 0), (–⅓, 0), (–1, 0) the solutions to 3x3 – 2x2 – 7x – 2 = 0 x = 2, –⅓, –1 the linear factors with real coefficients of 3x3 – 2x2 – 7x – 2 (x – 2)(x + ⅓)(x +1)

The factored form of the original polynomial is:

Number of Zeros A polynomial of degree n has at most n distinct zeros. Ex: State the maximum number of distinct real zeros of f: f(x) = 4x7 – 2x6 – 7x3 – 2x + 4 Answer: 7

Students Try!!! Find a polynomial function g of degree 4 such that the zeros of g are 0, -1, 2, -3, and g(3) = 288. Leave the polynomial in factored form. g(x) = a(x – 0)(x + 1)(x – 2)(x + 3) g(x) = a(x)(x + 1)(x – 2)(x + 3) 288 = a(3)(3 + 1)(3 – 2)(3 + 3) 288 = a(3)(4)(1)(6) 288 = 72a a = 4 g(x) = 4(x)(x + 1)(x – 2)(x + 3)