Descartes’ Rule of Signs Determines the possible nature of the zeros

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Presentation transcript:

7.5 Descartes’ Rule, Intermediate Value Theorem, Sum & Product of Zeros

Descartes’ Rule of Signs Determines the possible nature of the zeros We are going to learn more rules to help us better factor and find zeros of polynomial functions! Descartes’ Rule of Signs Determines the possible nature of the zeros  # of (+) or (–) real roots as well as imaginary roots How? • Count # of sign changes in polynomial this is # of (+) real roots or less by a multiple of 2 • Plug in (–x) for x & count # of sign changes again this is # of (–) real roots or less by a multiple of 2 • Enter in a table (+) (–) i *remembering each row must add to degree Ex 1) Determine the nature of zeros (+) (–) i + – + +  2 2 1 P(–x) = – – – +  1 1 2

Intermediate Value Theorem If P(x) is a polynomial function such that a < b and P(a) ≠ P(b), then P(x) takes on every value between P(a) and P(b) in the interval [a, b] *Special Case: If the values of P(x) change from (+) to (–) or from (–) to (+), there must be a 0 somewhere in between Ex 2) Use graphing calculator to find zeros to nearest hundredth 0.45, –4.03 Upper & Lower Bound Theorems: Using synthetic division on poly with real coeff & lead coeff is positive: (1) If c is a (+) real, & all #s in last row are non-negative, c is an upper bound (2) If c is a (–) real, & the #s in last row alternate in sign, c is a lower bound (0 can be written as +0 or –0) *Start at 0 & go up for upper bound, go down for lower bound (no zeros > c) (no zeros < c)

Ex 3) Determine the least integral upper bound and greatest integral lower bound for 2 –1 2 –1 –24 12 1 2 1 3 2 –22 Stop! negative 2 2 3 8 15 6 24  all (+) : upper bound –1 2 –3 5 –6 –18 Stop! no switch –2 2 –5 12 –25 26 –40  alternates : lower bound

Ex 4) Sketch a possible graph P(x) satisfying the following conditions: • 4th degree, leading coefficient 1 • 2 distinct negative real zeros & a positive real zero with mult 2 • greatest integral lower bound is –5 and least integral upper bound is 4 (edges going up) (crosses x-axis at 2 (–) values & is tangent at a (+) value) (zeros happen between –5 and 4) (Do on board) –5 4 (just one possible answer)

Putting it all together! Ex 5) Locate the real zeros of the polynomial between consecutive integers (+) (–) i nature of roots? + – – –  1 1 3 P(–x) = + – + –  3 1 1 2 *try for (+) first 1 0 –3 –6 –2 1 1 1 –2 –8 –10 2 1 2 1 –4 –10 here! 3 is upper bound! 3 1 3 6 12 34 1 0 –3 –6 –2 here! more? Between 2 & 3 and –1 & 0 –1 1 –1 –2 –4 2 lower bound so no more! –2 1 –2 1 –8 14

Homework #705 Pg 361 #1–15 odd, 16, 20–24, 28, 30, 36, 40, 42, 45, 47, 48, 51 Hint: For #48, read paragraph above