Today in Pre c alculus

1. Today in Precalculus • Go over homework • Notes: Remainder and Factor Theorems • Homework

2. Remainder Theorem If a polynomial f(x) is divided by x – k, then the remainder is r = f(k). - Find the remainder without doing synthetic division. Ex: Use the remainder theorem to determine the remainder when f(x) = x2 – 4x – 5 is divided by: • x – 3 k = 3 (3)2 – 4(3) – 5 = -8 • x + 2 k = -2 (-2)2 – 4(-2) – 5 = 7 • x – 5 k = 5 (5)2 – 4(5) – 5 = 0

3. Remainder Theorem Because the remainder in example c is zero, we know that x – 5 divides evenly into f(x). Therefore, 5 is a zero or x –intercept of the graph of f(x). And 5 is a solution or root of the equation f(x) = 0.

4. Fundamental Connections for Polynomial Functions For a polynomial function f and a real number k, the following statements are equivalent 1. x = k is a solution (or root) of the equation f(x) =0 2. k is an x-intercept of the graph of y = f(x) • k is a zero of the function f • x – k is a factor of f(x)

5. Factor Theorem A polynomial function f(x) has a factor x – k iff f(k) = 0.

6. Examples Use the factor theorem to determine if the first polynomial is a factor of the second polynomial. • x + 2; 4x3 – 2x2 + x – 5 k = -2 4(-2)3 – 2(-2)2 + (-2) – 5 = -32 – 8 – 2 – 5 = -47 ≠0, therefore, x + 2 is not a factor of 4x3 – 2x2 + x – 5 2. x + 2; x3 – 2x2 + 5x + 26 k = -2 (-2)3 – 2(-2)2 + 5(-2) + 26 = – 8 – 8 – 10 +26 =0 therefore, x + 2 is a factor of x3 – 2x2 + 5x + 26

7. Writing Polynomial Functions Example: Leading Coefficient: 3 Degree: 3 Zeros: -4, 3, -1 so factors are x + 4, x – 3, x + 1 3(x + 4)(x – 3)(x + 1) (3x + 12)(x2 – 2x – 3) 3x3 + 6x2 – 33x – 36

8. Writing Polynomial Functions Example: Leading Coefficient: 2 Degree: 3 Zeros: -3, -2, 5 so factors are x + 3, x + 2, x – 5 2(x + 3)(x + 2)(x – 5) (2x + 6)(x2 – 3x – 10) 2x3 – 38x – 60

9. Homework • Pg. 223: 13-24all, 27-30all