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Warm-up Find all the solutions over the complex numbers for this polynomial: f(x) = x4 – 2x3 + 5x2 – 8x + 4 {1, 1, 2i, 2i} Descartes Rule 2.5 Objectives I can use Descartes Rule to find the possible combinations of positive and negative real zeros I can write a polynomial in factor format Descartes’s Rule of Signs: If f(x) is a polynomial with real coefficients and a nonzero constant term, 1. The sign changes for f(x) tells the number of positive real zeros equal to the number of sign changes or less than that number by an even integer. 2. The sign changes on f(-x) tells the number of negative real zeros equal to the number sign changes or less than that number by an even integer. Example: Use Descartes’s Rule of Signs to determine the possible number of positive and negative real zeros of f(x) = 2x4 – 17x3 + 35x2 + 9x – 45. The polynomial has three variations in sign. + to – + to – f(x) = 2x4 – 17x3 + 35x2 + 9x – 45 – to + f(x) has either three positive real zeros or one positive real zero. f(– x) = 2(– x)4 – 17(– x)3 + 35(– x)2 + 9(– x) – 45 =2x4 + 17x3 + 35x2 – 9x – 45 f(x) has one negative real zero. One change in sign Find Numbers of Positive and Negative Zeros State the possible number of positive real zeros, negative real zeros, and imaginary zeros of p(x) = –x6 + 4x3 – 2x2 – x – 1. Since p(x) has degree 6, it has 6 zeros. However, some of them may be imaginary. Use Descartes’ Rule of Signs to determine the number and type of real zeros. Count the number of changes in sign for the coefficients of p(x). p(x) = –x6 + yes – to + 2 or 0 positive real zeros 4x3 – yes + to – 2x2 – no – to – x – no – to – 1 Find Numbers of Positive and Negative Zeros Since there are two sign changes, there are 2 or 0 positive real zeros. Find p(–x) and count the number of sign changes for its coefficients. p(–x) = –x6 no – to – 4x3 – 2x2 no – to – + yes – to + x – yes + to – Since there are two sign changes, there are 2 or 0 negative real zeros. 1 Find all of the zeros of f(x) = x3 – x2 + 2x + 4. Since f(x) has degree of 3, the function has at lost three zeros. To determine the possible number and type of real zeros, examine the number of sign changes in f(x) and f(–x). f(x) = x3 – x2 yes f(–x) = –x3 – no + 2x yes x2 – no + 4 2 or 0 positive real zeros no 2x + yes 4 1 negative real zero Descarte’s Rule of Signs Find how many positive and negative roots there are in f(x). f(x) = 3x4 + 2x3 + 1 0 positive roots 2 or 0 negative roots Descarte’s Rule of Signs Find how many positive and negative roots there are in f(x). f(x) = 3x3 + 2x2 - x + 3 2 or 0 positive roots 1 negative root Descarte’s Rule of Signs How can this rule help us with the rational root theorem? It helps us guess which possible root to try. f(x) = x – x + 2x + 4. 3 2 x 1 i 3, 1 Possible roots are ±1, ±2, ±4 There are 3 zeros, we know there are 2 or 0 positive and 1 negative -4, -2, -1, 1, 2, 4 Use Synthetic Division to find the roots Use Quadratic Formula to find remaining solutions 211111- 1- 1- 1222 444 We need either 2 or 0 positive, so 4 cannot be a zero 21- 1 202 82- 4 1 111 0- 242 4126 0 -1 works, so it is our 1 negative zero. The other two have to be imaginary. 2 4 4(1)( 4) 2 12 2 2i 3 1 i 3 2(1) 2 2 x 1 2i, 1, 3 Solve 0 x 4 2 x 2 16 x 15 There are 4 zeros, We know there are 1 positive and 3 or 1 negative (-3x)-1, 1, (3,x)5, 152( x) -15,f-5, 4 Possible 16(are x±1, ) 15 ±3, ±5, ±15 2 roots 35 11 00 --22 --16 16 --15 15 Now try negatives 35 25 9 115 21 15 495 11 35 23 7 99 5 480 0 13 11 33 77 55 - 13 -02 -- 21 5 1 02 57 |-016 3 is Allapositive solution.soThis 5 isisanour 1upper positive bound zero -1 Alternates is our 1 negative. between Use positive the quadratic and negative to find the so 3remaining is a lower2 bound zeros 2 4 4(1)(5) 2 16 2 4i 1 2i 2(1) 2 2 Linear Factorization If we know the solutions, we can work backwards and find the Linear Factorization Example:Given the solutions to a polynomial are:{-2, 3, 6, 2+3i, 2-3i} write the polynomial as a product of its factors f(x) = (x+2)(x-3)(x-6)(x-(2+3i))(x-(2-3i)) Also Note: If we know all the factors and multiply them together, we get the polynomial function. Homework WS 4-3