pencil, red pen, highlighter, GP notebook

1. total: pencil, red pen, highlighter, GP notebook M3D16 Have out: Bellwork: Determine Solve the problem using both LONG DIVISION and SYNTHETIC DIVISION. Show both methods. P(x) = x3 + 4x2 – 7x – 14 and d(x) = x – 2

2. P(x) = x3 + 4x2 – 7x – 14 and d(x) = x – 2 Long Division: + 1 +1 +1 + – + 1 + 2 – + – + +1 work

3. total: P(x) = x3 + 4x2 – 7x – 14 and d(x) = x – 2 Synthetic Division: +1 1 4 –7 –14 +10 +2 +12 +1 +1 1 6 5 2 +1 –4 • • • +1 +2

4. Example #1: Say we are given P(x) = x3 – 5x2 + 9x - 5, how can we find the zeros? Does it factor? ??? Nope, group factoring does not work, nor will any other type of factoring. Then what? Where do we start looking for zeros of P(x)?

5. Rational Zero Theorem For any polynomial , every rational ______ of P(x) has the form , where: zero factor of a0 constant p = ____________ (a0 is a _________) factor of an leading coefficient q = ____________ (an is a _________________)

6. 1 Back to p q Possible rational zeros ±1, ±5 factors of –5 make combinations… ±1 factors of 1 All of these answers are possible zeros for P(x), but not all of them will be zeros. We just have to pick the ones that will work in P(x). 1 Here’s a start: what is the easiest number you can test?

7. 5 –4 Possible rational zeros ±1, ±5 1 Try x = ___ 1 –5 9 –5 Yes! On our first try we found a zero since there is no remainder!!! +5 +1 –4 1 –4 5 1  • • • Can the quadratic be factored using a diamond problem? No

8. Find the remaining zeros of P(x) by completing the square on ___________ = 0. zeros of P(x) = _______

9. 1 Example #2: Find all the zeros of p q Possible rational zeros ±1, ±2, ±4, ±8, make combinations… factors of –8 ±1 factors of 1 1 Start with x = ____ What a bummer! We have a remainder, so x = 1 is NOT a zero. We need to go back and try another possible zero. –8 1 0 –11 –18 –28 –10 +1 +1 What is another easier number to test? 1 1 –10 1 –28 –36 • • • • 2, you say? Let’s try it…

10. 1 Example #2: Find all the zeros of p q Possible rational zeros ±1, ±2, ±4, ±8, make combinations… factors of –8 ±1 factors of 1 2 Try x = ____ Not again! We have a remainder, so x = 2 is also NOT a zero. –8 1 0 –11 –18 –64 –14 +2 +4 What should we try this time? 1 2 –7 2 –32 –72 A negative number? • • • • Let’s try –1.

11. 1 Example #2: Find all the zeros of p q Possible rational zeros ±1, ±2, ±4, ±8, make combinations… factors of –8 ±1 factors of 1 –1 Try x = ____ Success!!! Therefore, (x + 1) is a factor of P(x). –8 1 0 –11 –18 +8 +10 –1 +1 1 –1 –10 –1 –8  • • • • We have

12. We need to keep breaking down x3 – x2 – 10x – 8, but group factoring will not work. We must use the Rational Zero Theorem again to investigate other possible rational zeros. 1 Possible rational zeros of ________________ = 0 p q factors of –8 ±1, ±2, ±4, ±8, factors of 1 ±1 Hint: Try x = –1 again. –1 Try x = ____ 1 –1 –10 –8 Factor! +8 –1 +2 1 –2 –8 –1  • • • –1, –2, –1, 4 zeros of P(x): x = ____________

13. Example #3: Find all the zeros of p q Possible rational zeros ±1, ±2, ±3, ±4, ±6, ±12 factors of 12 ±1, ±2 factors of 2 = ±1, ±2, ±3, ±4, ±6, ±12 4 There is a remainder, so x = 4 is not a zero. Try x = ____ –25 12 2 1 Looking at the above list, there are many possible zeros. Is there a way to efficiently pick a zero without completely guessing? +44 +36 +8 2 9 11 4 56 • • • Some theorems might help…

14. 1 –25 12 2 +44 +8 +36 2 11 9 4 56 Upper Bound Theorem all When dividing P(x) by (x – k), and k > 0, and _____ entries in the last row are ≥ 0, then k is an ______ ________ for the _____ zeros of P(x). For example, ___ _____ of this P(x) is ________ than ____. upper bound real no zero greater 4 Look at the last row. Since all the numbers are positive, then 4 is an upper bound. We can eliminate all possible zeros that are greater than or equal to 4.

15. Example #3: Find all the zeros of p q –6 Try x = ____ Since there is a remainder, x = –6 is not a zero. 2 –25 12 1 –246 –12 +66 This may not seem like useful information, but x = –6 shows us another bound for the possible rational zeros. 2 –11 41 –6 –234 • • •

16. 2 –25 12 1 –246 –12 +66 2 –11 41 –6 –234 Lower Bound Theorem When dividing P(x) by (x – k), and k < 0, and the entries in the last row are _________ between ________ and ________ (FYI, _____ counts as ________ or ________), then k is a ______ _______ for the _____ zeros of P(x). For example, ___ _____ of this P(x) is _____ than ___. positive negative alternates positive negative zero lower bound real –6 no zero less Look at the last row. Since all the numbers alternate between positive and negative, then –6 is a lower bound. We can eliminate all possible zeros that are less than or equal to –6. Pick a possible zero. ½, you say? Let’s try it.

17. Example #3: Find all the zeros of p q ½ Try x = ____ Success! x = ½ is a zero. 2 –25 12 1 –12 +1 +1 2 –24 2 ½  • • • Let’s graph y = P(x). –4, ½, 3 zeros of P(x): x = ________

18. y x (0, 12) –4, ½, 3 zeros of P(x): x = ________ Before we graph, go through the check list: (3, 0) (–4, 0) (½ , 0)  What’s the degree? 3  What’s the leading coefficient? +2  What’s the end behavior?  Any double zeros? No

19. Complete the practice worksheet

20. Mixed Practice 1. Find all zeros of P(x). by first listing all possible rational zeros. Sketch a graph of y = P(x). a) x = – 6, –1, 1 b) x = –2, –2, 4