1. 6.3 Dividing Polynomials

2. Warm Up Without a calculator, divide the following Solution: 49251

3. This long division technique can also be used to divide polynomials

4. POLYNOMIALS – DIVIDING EX – Long division (5x³ -13x² +10x -8) / (x-2) 5x³ - 13x² + 10x - 8x - 2 5x² 5x³ - 10x²-() -3x²+ 10x - 3x -3x² + 6x - () 4x - 8 4x - 8 - () + 4 0 R 0

5. So in other words… 5x³ -13x² +10x - 8 x-2 = 5x² -3x + 4 (5x² -3x + 4) (x-2) = 5x³ -13x² +10x - 8 OR

6. POLYNOMIALS – DIVIDING EX#2 – Long division (x² +3x -12) / (x+6) x² + 3x - 12x - 3 x x² - 3x-() 6x-12 + 6 6x - 18 - () 6 R 6

7. (2x² -19x + 8) / (x-8) 2x² - 19x + 8x - 8 Let’s Try One

8. (2x² -19x + 8) / (x-8) 2x² - 19x + 8x - 8 Let’s Try One

9. EX – Synthetic division (5x³ -13x² +10x -8) / (x-2) 2 5 -13 10 -8 5 10 -3 -6 4 8 0 5x² -3x + 4 Opposite of number in divisor

10. EX – Synthetic division (3x³ -4x² +2x -1) / (x+1) -1 3 -4 2 -1 3 -3 -7 +7 9 -9 -10 3x 2 -7x + 9 R-10 Opposite of number in divisor

11. Let’s Try One (x³ -13x +12) / (x+4)

12. EX – Synthetic division (x³ -13x +12) / (x+4) Opposite of number in divisor

13. A Couple of Notes Use synthetic division when the coefficient in front of x is 1 (x- 2) (2x-3) YES NO To test so see if a binomial is a factor, you want to see if you get a remainder of zero. If yes, it is a factor. If you get a remainder, the answer is no.

14. (2x² -19x + 8) / (x-8) 2x² - 19x + 8x - 8 From this example, x-8 IS a factor because the remainder is zero

15. In this case, x-3 is not a factor because there was a remainder of 6 x² + 3x - 12x - 3 x x² - 3x-() 6x-12 + 6 6x - 18 - () 6 R 6