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(These will be on tomorrow’s quiz!)
pencil, red pen, highlighter, GP notebook U7D12 Have out: Bellwork: Determine Solve the problem using both LONG DIVISION and SYNTHETIC DIVISION. Show both methods. (These will be on tomorrow’s quiz!) P(x) = x3 + 4x2 – 7x – 14 and d(x) = x – 2 total:
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P(x) = x3 + 4x2 – 7x – 14 and d(x) = x – 2
Long Division: + 1 +1 +1 – + + 1 + 2 – + – + +1 work
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P(x) = x3 + 4x2 – 7x – 14 and d(x) = x – 2
Synthetic Division: +1 1 4 –7 –14 +2 +12 +10 +1 +1 2 1 6 5 –4 +1 • • • +1 +2 total:
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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)?
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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 _________________)
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Possible rational zeros
Back to 1 p q Possible rational zeros factors of –5 ±1, ±5 make combinations… factors of 1 ±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?
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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!!! +1 –4 +5 1 1 –4 5 • • • Can the quadratic be factored using a diamond problem? 5 –4 No
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Find the remaining zeros of P(x) by completing the square on ___________ = 0.
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Possible rational zeros
Example #2: Find all the zeros of 1 p q Possible rational zeros factors of –8 ±1, ±2, ±4, ±8, make combinations… factors of 1 ±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. 1 –11 –18 –8 +1 +1 –10 –28 What is another easier number to test? 1 1 1 –10 –28 –36 • • • • 2, you say? Let’s try it…
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Possible rational zeros
Example #2: Find all the zeros of 1 p q Possible rational zeros factors of –8 ±1, ±2, ±4, ±8, make combinations… factors of 1 ±1 2 Try x = ____ Not again! We have a remainder, so x = 2 is also NOT a zero. 1 –11 –18 –8 +2 +4 –14 –64 What should we try this time? 2 1 2 –7 –32 –72 A negative number? • • • • Let’s try –1.
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Possible rational zeros
Example #2: Find all the zeros of 1 p q Possible rational zeros factors of –8 ±1, ±2, ±4, ±8, make combinations… factors of 1 ±1 –1 Try x = ____ Success!!! Therefore, (x + 1) is a factor of P(x). 1 –11 –18 –8 –1 +1 +10 +8 –1 1 –1 –10 –8 • • • • We have
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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. Possible rational zeros of ________________ = 0 1 q p factors of –8 ±1, ±2, ±4, ±8, factors of 1 ±1 Hint: Try x = –1 again. –1 Try x = ____ 1 –1 –10 –8 Factor! –1 +2 +8 –1 1 –2 –8 • • • zeros of P(x): x = ____________ –2, –1, –1, 4
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Possible rational zeros
Example #3: Find all the zeros of p q Possible rational zeros factors of 12 ±1, ±2, ±3, ±4, ±6, ±12 factors of 2 ±1, ±2 = ±1, ±2, ±3, ±4, ±6, ±12 4 Try x = ____ There is a remainder, so x = 4 is not a zero. 2 1 –25 12 Looking at the above list, there are many possible zeros. Is there a way to efficiently pick a zero without completely guessing? +8 +36 +44 4 2 9 11 56 • • • Some theorems might help…
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Upper Bound Theorem 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 ____. all upper bound real no zero greater 4 56 2 1 –25 12 4 +8 9 +36 11 +44 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.
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Example #3: Find all the zeros of p q
Try x = ____ –6 Since there is a remainder, x = –6 is not a zero. 2 1 –25 12 –12 +66 –246 This may not seem like useful information, but x = –6 shows us another bound for the possible rational zeros. –6 2 –11 41 –234 • • •
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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 ___. alternates positive negative zero positive negative lower bound real no zero less –6 Look at the last row. Since all the numbers alternate between positive and negative, then –6 is a lower bound. –234 2 1 –25 12 –6 –12 –11 +66 41 –246 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.
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Example #3: Find all the zeros of p q Try x = ____ ½ 2 1 –25 12 +1
Success! x = ½ is a zero. 2 1 –25 12 +1 +1 –12 2 2 –24 • • • Let’s graph y = P(x). zeros of P(x): x = ________ ½, –4, 3
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zeros of P(x): x = ________ ½, –4, 3
y (0, 12) zeros of P(x): x = ________ ½, –4, 3 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
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Complete the practice worksheet
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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
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