Sullivan Algebra and Trigonometry: Section 5.4 Objectives • Utilize the Conjugate Pairs Theorem to Find the Complex Zeros of a Polynomial • Find a Polynomial Function with Specified Zeros • Find the Complex Zeros of a Polynomial A variable in the complex number system is referred to as a complex variable. A complex polynomial function f degree n is a complex function of the form f ( x ) an x an 1 x n n 1 a1 x a0 A complex number r is called a (complex) zero of a complex function f if f (r) = 0. Note that real numbers are also complex numbers in the form a + bi where b = 0. So, this definition of a complex polynomial function is a generalization of what was previously introduced. Fundamental Theorem of Algebra Every complex polynomial function f (x) of degree n > 1 has at least one complex zero. Theorem Every complex polynomial function f (x) of degree n > 1 can be factored into n linear factors (not necessarily distinct) of the form f ( x ) an ( x r1 )( x r2 ) ( x rn ) where an , r1 , r2 , , rn are complex numbers. Conjugate Pairs Theorem Let f (x) be a complex polynomial whose coefficients are real numbers. If r = a + bi is a zero of f, then the complex conjugate r a bi is also a zero of f. Corollary A complex polynomial f of odd degree with real coefficients has at least one real zero. Find a polynomial f of degree 4 whose coefficients are real and has zeros 0, -2 and 1 - 3i. Graph f to verify the solution. f ( x ) a ( x 0)( x ( 2))( x (1 3i ))( x (1 3i )) f ( x) x( x 2)( x 2 (1 3i ) x (1 3i ) x (1 3i )(1 3i ) f ( x) ( x 2 2 x)( x 2 x 3ix x 3ix 1 3i 3i 9i 2 ) f ( x ) ( x 2 2 x )( x 2 2 x 10) f ( x ) x 2 x 10x 2 x 4 x 20x 4 3 f ( x ) x 6 x 20 x 4 2 2 3 2 Find the complex zeros of the polynomial f ( x) 2 x 3x 40 x 137 x 58. 4 3 2 There are 4 complex zeros. Using Descartes’ Rule of Sign, there are two, or no, positive real zeros. f ( x ) 2 x 4 3x 3 40x 2 137 x 58 Using Descartes’ Rule of Sign, there are two, or no, negative real zeros. Now, list all possible rational zeros p/q by factoring the first and last coefficients of the function. f ( x) 2 x 4 3x 3 40 x 2 137 x 58. p: 1, 2, 29, 58 q: 1, 2 p 1 29 : 1, 2, 29, 58, , q 2 2 Now, begin testing each potential zero using synthetic division. If a potential zero k is in fact a zero, then x - k divides into f (remainder will be zero) and is a factor of f. f ( x) 2 x 3x 40 x 137 x 58. 4 3 Test k = -2 - 2 2 2 -3 -4 -7 2 40 137 58 14 -108 - 58 54 29 0 Thus, -2 is a zero of f and x + 2 is a factor of f. f ( x ) ( x 2)(2 x 3 7 x 2 54 x 29) -1 -7 54 29 -1 4 - 29 2 -8 58 0 Thus, -1/2 is a zero of f and x + 1/2 is a factor of f. f ( x ) x 2 x 1 2 x 2 8 x 58 2 Test k = -1/2 2 2 f ( x ) x 2 x 1 2 f ( x ) 2x 2 x 1 2 2 x 8x 58 2 x 2 4 x 29 Now, find the complex zeros of the quadratic factor. ( 4) ( 4) 2 4(1)(29) x 2(1) 4 100 2 4 10i 2 2 5i Zeros: - 2, - 1 , 2 5i , 2 + 5i 2