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Zeros of Polynomials 2.5 Properties of Polynomial Equations A polynomial of degree n, has n roots (counting multiple roots separately) If a + bi is a root, then a – bi is also a root. Complex roots occur in conjugate pairs! Descartes’ Rule of Signs 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) = –(–x)6 + no – to – 4(–x)3 – 2(–x)2 – no – to – yes – to + (–x) – yes + to – Since there are two sign changes, there are 2 or 0 negative real zeros. Make a chart of possible combinations. 1 Find Numbers of Positive and Negative Zeros Answer: Find all of the zeros of f(x) = x3 – x2 + 2x + 4. Since f(x) has degree of 3, the function has 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 Rational Zero Theorem Factors of a 0 Possible Zeros Factors of a n Identify Possible Zeros A. List all of the possible rational zeros of f(x) = 3x4 – x3 + 4. Answer: B. List all of the possible rational zeros of f(x) = x3 + 3x + 24. <YU TRY> A. B. C. D. Use the Factor Theorem Show that x – 3 is a factor of x3 + 4x2 – 15x – 18. Then find the remaining factors of the polynomial. The binomial x – 3 is a factor of the polynomial if 3 is a zero of the related polynomial function. Use the factor theorem and synthetic division. 1 4 –15 3 21 1 7 6 –18 18 0 Use the Factor Theorem Since the remainder is 0, (x – 3) is a factor of the polynomial. The polynomial x3 + 4x2 – 15x –18 can be factored as (x – 3)(x2 + 7x + 6). The polynomial x2 + 7x + 6 is the depressed polynomial. Check to see if this polynomial can be factored. x2 + 7x + 6 = (x + 6)(x + 1) Answer: Factor the trinomial. So, x3 + 4x2 – 15x – 18 = (x – 3)(x + 6)(x + 1). Example 0 x 4 6 x 2 8 x 24 Solve Possible Zeros : 1, 2, 3, 4, 6, 8, 12, 24 2 1 0 -6 2 4 -8 24 2 1 2 -2 - 12 2 8 12 1 4 6 0 - 4 - 24 1 2 - 2 - 12 0 x 4 6 x 3 8 x 24 x 2 4 x 6 x 2 Repeated zero at x = 2 2 Upper and Lower Bounds Suppose f(x) is divided by x – c using synthetic division If c > 0 and each number in the last row is either positive or zero, then c is an upper bound for real zeros If c < 0 and each number in the last row are alternatively positive or negative (zero counts as both), then c is a lower bound Solve 0 x 4 6 x 3 22 x 2 30 x 13 Possible roots are ±1, ±13 The degree is 4, so there are 4 roots! 1 1 - 6 22 Use Synthetic Division to find the roots 1 -5 - 30 13 17 - 13 1 - 5 17 - 13 0 4 16 4(1)(13) 4 36 4 6i 2 3i 2(1) 2 2 Multiplicity of 2 x 1, 1, 2 3i 1 1 - 5 17 - 13 1 13 -4 1 - 4 13 0 x 2 4 x 13 Use Quadratic Formula to find remaining solutions 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- 1202 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 75 |-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 Theorem Find a fourth degree polynomial with real coefficients that has 2, -2 and i as zeros and such that f(3) = -150 f ( x) an x 2x 2x i x i f ( x) a x 3x 4 150 a (3) 3(3) 4 f ( x ) an x 2 4 x 2 1 4 2 n 4 2 n 3 an Foil Foil Substitute f(3) = -150 Solve for a Substitute back in equation f ( x) 3 x 4 3x 2 4 3x 4 9 x 2 12 Linear Factorization Theorem An nth – degree polynomial can be expressed as the product of a nonzero constant and n linear factors Completely factor x 4 3x 2 28 x 2 7 x 7 x2 4 factored over the rationals x 7 x 4 factored over the reals x 7 x 7 x 2i x - 2i completely factored 2