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Chapter 4: Polynomial and Rational Functions Section 4-5: Locating Zeros of a polynomial function The Location Principle • Suppose y=f (x) represents a polynomial function with real coefficients. If a and b are two numbers with f (a) negative and f (b) positive, the function has at least one real zero between a and b. • When we set up a table of synthetic division, a change in signs in the remainder indicates that there is a zero between the two integers. Example • Determine between which consecutive integers the real zeros of are located. F(x)=12x3-20x2-x+6 • There are three complex zeros for this function. According to Descartes’ Rule of Signs, there are two or zero positive real roots and one negative root. Using synthetic division: So the zeros according to the location Principle, will be between -1 and 0, 0 and 1, 1 and 2 r 12 -20 -1 6 -3 12 -56 -2 12 -44 -1 12 -32 - 0 12 -20 -1 + 1 12 -8 -9 -3 2 12 4 7 20 3 12 16 4 12 + + Upper and Lower Bound • The upper and lower bound theorem helps you narrow down the possibilities of zeros from p /q without graphing, and helps you make sure you found all of them. Upper Bound Theorem • Suppose c is a positive real number and P (x) is divided by x - c. If the resulting quotient and remainder have no change is sign, then P (x) has no real zero greater than c. Thus c is an upper bound of the zeros of P (x). (Zero coefficients are ignored when counting sign changes). • What you are looking for: In the table using synthetic division all of the signs across the row are the same. Lower Bound Theorem • If c is an upper bound of the zeros of P (-x), then –c is a lower bound of the zeros of P (x). Example • Use the Upper Bound Theorem to find an integral upper bound and the Lower Bound Theorem to find an integral lower bound of the zeros of F(x)=6x3-7x2-14x+15 • The rational root theorem tells us that ±15/2, ±1, ±3, ±5, ± 15, ±1/2, ±1/3, ±1/6, ±3/2, ±5/2, ±5/3, ±5/6, r 6 1/2 6 -4 5/6 6 1 6 -1 3 6 -7 -14 15 ±15/6 -2 11 19 72 No change in signs 3 is the upper bound Finding the Lower Bound r • Let’s2 look at f (-x)=6(-x)3- 1 7(-x) -14(-x)+15= -6x3-7x2+14x+15 2 • I do not have any r>0 3 (negative) because a negative r x -6 would give 4 me a positive 5 • 2 is an upper bound of • f (-x), so -2 is a lower 6 bound of f (x) 7 8 9 -6 -6 -6 -6 -6 -6 -6 -6 -6 -6 -7 14 15 -13 +1 -19 -24 -33 HW#29 • Section 4-5 • Pp 240-242 • #13-25 odd, 41-44 all