1. The Remainder and Factor Theorems 6.5 p. 352

2. When you divide a Polynomial f(x) by a divisor d(x), you get a quotient polynomial q(x) with a remainder r(x) written: f(x) = q(x) + r(x) d(x) d(x)

3. The degree of the remainder must be less than the degree of the divisor!

4. Polynomial Long Division: You write the division problem in the same format you would use for numbers. If a term is missing in standard form …fill it in with a 0 coefficient. Example: 2x 4 + 3x 3 + 5x – 1 = x 2 – 2x + 2

5. 2x 4 = 2x 2 x 2 2x 2

6. +4x 2 -4x 3 2x 4 -( ) - 4x 2 7x 3 +5x 7x 3 = 7x x 2 +7x 7x 3 - 14x 2 +14x-( ) 10x 2 - 9x +10 10x 2 - 20x +20-( ) 11x - 21 remainder

7. The answer is written: 2x 2 + 7x + 10 + 11x – 21 x 2 – 2x + 2 Quotient + Remainder over divisor

8. Now you try one! y 4 + 2y 2 – y + 5 = y 2 – y + 1 Answer: y 2 + y + 2 + 3 y 2 – y + 1

9. Remainder Theorem: If a polynomial f(x) is divisible by (x – k), then the remainder is r = f(k). Now you will use synthetic division (like synthetic substitution) f(x)= 3x 3 – 2x 2 + 2x – 5 Divide by x - 2

10. f(x)= 3x 3 – 2x 2 + 2x – 5 Divide by x - 2 Long division results in ?...... 3x 2 + 4x + 10 + 15 x – 2 Synthetic Division: f(2) = 3-22-5 2 3 6 4 8 10 20 15 Which gives you: 3x 2 + 4x + 10 + 15 x-2

11. Synthetic Division Practice 1 Divide x 3 + 2x 2 – 6x -9 by (a) x-2 (b) x+3 (a) x-2 12-6-9 2 1 2 4 8 2 4 -5 Which is x 2 + 4x + 2 + -5 x-2

12. Synthetic Division Practice cont. (b) x+3 12-6-9 -3 1 -3 3 -3 9 0 x 2 – x - 3

13. Factor Theorem: A polynomial f(x) has factor x-k iff f(k)=0 note that k is a ZERO of the function because f(k)=0

14. Factoring a polynomial Factor f(x) = 2x 3 + 11x 2 + 18x + 9 Given f(-3)=0 Since f(-3)=0 x-(-3) or x+3 is a factor So use synthetic division to find the others!!

15. Factoring a polynomial cont. 211189 -3 2 -6 5 -15 3 -9 0 (x + 3)(2x 2 + 5x + 3) So…. 2x 3 + 11x 2 + 18x + 9 factors to: Now keep factoring (bustin ‘da ‘b’) gives you: (x+3)(2x+3)(x+1)

16. Your turn! Factor f(x)= 3x 3 + 13x 2 + 2x -8 given f(-4)=0 (x + 1)(3x – 2)(x + 4)

17. Finding the zeros of a polynomial function f(x) = x 3 – 2x 2 – 9x +18. One zero of f(x) is x=2 Find the others! Use synthetic div. to reduce the degree of the polynomial function and factor completely. (x-2)(x 2 -9) = (x-2)(x+3)(x-3) Therefore, the zeros are x=2,3,-3!!!

18. Your turn! f(x) = x 3 + 6x 2 + 3x -10 X=-5 is one zero, find the others! The zeros are x=2,-1,-5 Because the factors are (x-2)(x+1)(x+5)

19. Assignment