Complex Conjugation and Polynomial Factorization

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1 Comlex Conjugation and Polynomial Factorization Dave L. Renfro Summer 2004 Central Michigan University I. The Remainder Theorem Let P (x) be a olynomial with comlex coe cients 1 and r be a comlex number. remainder when P (x) is divided by x r is P (r). The Examle 1: If P (x) = 2ix 2 6x + (4 i) and r = 4i, then by long division or by synthetic division you should be able to verify that 2ix 2 6x + (4 i) x 4i = ix i x 4i Also, you should be able to verify by direct substitution that P (4i) = 4 59i. Proof: If we write the result of dividing P (x) by x r in uotient + remainder form, we have P (x) = (x r) Q (x) + R, where Q (x) is the uotient and R (a constant 2 ) is the remainder. Now lug x = r into both sides. This gives P (r) = R, which is what we were to rove. II. The Factor Theorem Let P (x) be a olynomial with comlex coe cients suose r is a comlex number such that P (r) = 0. Then x r is a factor of P (x). You should be able to see that this is a secial case of the Remainder Theorem. Examle 2: If P (x) = x 12x x 148 and r = 2, then it is easy to verify that P (2) = 0. Hence, x 2 must be a factor of P (x). Using long division or synthetic division, you should be able to nd that P (x) = (x 2) x 2 10x Note that armed with this factorization, you can nd the other two zeros of this olynomial: x = 5 7i. Examle : Without rewriting, we can tell that x (i.e. x 0) is a factor of (x + y + z) (xy + yz + xz) (x + y) (y + z) (x + z) since utting x = 0 in this exression gives zero. [Suose y = 5 and z = 2. Then the exression becomes (x + ) (5x 10 2x) (x + 5) () (x 2), and this euals zero when x = 0. Thus, x is a factor of (x + y + z) (xy + yz + xz) (x + y) (y + z) (x + z) when y = 5 and z = 2. In the same way, x is a factor of this exression for any two other choices of y and z, since no matter what y and z are, the exression algebraically reduces to zero when x = 0. exression no matter what y and z are.] Hence, x is a factor of this 1 Kee in mind that every real number is also a comlex number. Real numbers are just comlex numbers that have the form a + 0 i. 2 Do you know why R will be a constant?

2 Examle 4: We can see that m + n is a factor of m 4 4m n + 2m 2 n 2 + 5mn 2n 4 since lugging n in for m in this exression gives zero. If you use long division or synthetic division, you ll nd the uotient after dividing by m + n is m 5m 2 n + 7mn 2 2n. Examle 5: Suose we want to factor (x + y + z) x + y + z Since x = y makes this zero, x+y is a factor. In the same way (or use symmetry), we nd that y + z and x + z are factors. Hence, (x + y) (y + z) (x + z) is a factor. Since this roduct has the same degree as the exression we re trying to factor, it follows that (x + y + z) x + y + z = k (x + y) (y + z) (x + z) for some constant k. 4 To nd k, we simly lug in three numbers and solve for k. For examle, x = 0, y = 1, and z = 2 gives us ( ) = k (0 + 1) (1 + 2) (0 + 2). Therefore, 18 = 6k, and hence k =. Examle 6: By using your calculator it would aear that = 1, but how could we rove this? 5 One way is to verify by hand that this number is a solution to x + 15x 16 = 0 and then use the fact that x = 1 is the only real solution of x + 15x 16 = 0, since x + 15x 16 = (x 1) x 2 + x I challenge you to nd another way to rove this number is 1! III. Using the Factor Theorem to solve euations 1. x + 2x 2 20x 16 = 0, given that x = 4 is a solution. Answer: x = 4; x 6x 2 = 90x + 148, given that x = 2 is a solution. Answer: x = 2; x 4 10x +70x 2 +40x 296 = 0, given that x = 2 are solutions. Answer: x = 2; 5 7i. 4. x 5 x 4 +x = 10x 10x 2 +1, given that x = 1 is a solution. Answer: x = 1; This is because no two of x + y, y + z, and x + z share any factors. Note that (x y) 2 and (x y) are factors of (x y) 4, but their roduct isn t. 4 Because dividing (x + y + z) x + y + z by (x + y) (y + z) (x + z) must give an algebraic exression of zero degree, and only constants have zero degree. 5 Comare this to, for examle, to the task of showing that is eual to 2, where exanding and simlifying the suare of this exression will work. 6 Verifying that this number is a solution to x + 15x 16 = 0 is not as tedious as you might think. Let x = Using the identity (a b) = a a 2 b + ab 2 b and short cuts such as = = , it doesn t take much to show that x = But this is just , which is clearly eual to 16 15x. Thus, x = 16 15x, and so x + 15x 16 = 0.

3 IV. Finding a olynomial with seci ed real zeros If a olynomial has zeros r 1 and r 2, then (x r 1 ) (x r 2 ) is a olynomial that has r 1 and r 2 as zeros. In fact, if P (x) is any olynomial, then P (x) (x r 1 ) (x r 2 ) will also be a olynomial that has r 1 and r 2 as zeros, although the resence of P (x) will likely generate additional zeros. Examle 7: x 2 x is a olynomial with 2 and as zeros. We can nd a olynomial with integer coe cients having 2 and as zeros if we air each of these factors with their conjugates in the following way. Note that what I m doing amounts to choosing P (x) = x + 2 x +. x 2 x + 2 x x + = x 2 2 x 2 = x 4 5x 2 +6 A more systematic way to obtain this nice olynomial is to nd nice olynomials for 2 and searately and then multily these two nice olynomials. Begin by utting x = 2, then suare both sides, and then subtract 2 from both sides to get a 0 on one side. The other side will be a olynomial (seci cally, x 2 2) that has 2 as a zero. If you do the same thing for x =, you ll get x 2. Examle 8: Here s how to nd a olynomial with integer coe cients that has 2 + as a zero. Let x = 2 + and suare both sides. This gives x 2 = Now isolate 2 6 and suare again: x = The olynomial we want will be the olynomial that shows u when we get a zero on one side of the euation, which is x 4 10x Since this is a 4 th degree olynomial, there should be four zeros, counting multilicity. As you can verify by hand calculation, each of the four sign choices in 2 give a zero of this olynomial. On the other hand, if you solve this by using the uadratic formula (the euation is uadratic in x 2 ), you ll get x = Note this is consistent with the third sentence in this examle, where I wrote x 2 = Thus, must eual 2 +, must eual 2 +, etc. In older college algebra texts (at least 50 years old) you can often nd a discussion of techniues for rewriting exressions of the form A + B as a sum or di erence of suare roots. This was useful because, for instance, 2 + tends to be easier to work with than , lus it s a lot easier to obtain a numerical aroximation for 2 + than it is for In the case of 2 +, you simly looked u the values of 2 and in a table and added the aroximate values by hand. A more exotic examle is being eual to

4 V. Problems: Finding euations with seci ed real zeros Find olynomials with integer coe cients whose zeros include the given numbers. You may leave your answers in factored form A olynomial whose zeros include 2 One answer: (x 2) A olynomial whose zeros include 2 One answer: (x 2) 2 5 (x + 4) and 4.. A olynomial whose zeros include 2 5, 4, and 2. One answer: (x 2) 2 5 (x + 4) x A olynomial whose zeros include 2 +. One answer: x + 9x x A olynomial whose zeros include h One answer: x i VI. Comlex zeros of olynomials with real coe cients If a olynomial has real coe cients, then its zeros occur in comlex conjugate airs. 8 I went through the details of this in class. The basic idea is that comlex conjugation distributes through addition and multilication (i.e. (z 1 + z 2 ) = z 1 + z 2 and (z 1 z 2 ) = z 1 z 2 ), and from this you can show that P (z) = P (z) if P is a olynomial with real coe cients. Therefore, if P (c) = 0 (i.e. c is a zero of P ) then, by taking the comlex conjugate of both sides, we get P (c) = 0, which imlies P (c) = 0 (i.e. c is a zero). 9 7 Hint for V(4): Beginning with x = 2 +, isolate the cube root, then cube both sides, then simlify all numerical suare roots so that they will be multiles of, then isolate all terms involving on one side of the euation, then suare both sides, then get a zero on one side. Since I have a little extra room on this age, here s a numerical marvel for your amusement: sin 17 s = r More generally, we have the following result. Let P be a non constant olynomial with comlex coe - cients. Then the zeros of P occur in comlex conjugate airs if and only if there exists a line L in the comlex lane such that 0 2 L and all the coe cients of P belong to L. This is Monthly Problem E 11 (roosed by Murray R. Siegel), whose solution (by F. D. Parker) is given in American Mathematical Monthly 62 #4 (Aril 1955), : If the coe cients lie on a common line, = 1, through the origin, then writing the coe cients in olar form roduces a factor e i 1. When this factor is divided out, the coe cients are real and the imaginary [non real] roots are in conjugate airs. If the imaginary roots occur in conjugate airs, then an euation with real coe cients can be formed with those roots. Any other euation with the same roots must di er by only a multilicative constant. Any such constant will lace the new coe cients on a common line through the origin. 9 Here s an interesting alternative method given by D. Trifan in Comlex roots of an integral rational euation, Amer. Math. Monthly 61 (1954), If P (a + bi) = 0, then we can factor P (x) as

5 Alication: If P (x) is an odd degree olynomial with real coe cients, then P (x) has at least one real zero. A calculus way to rove this same result is to observe that olynomials are continuous on every interval, and hence the Intermediate Value Theorem holds on every interval, which imlies 10 exists a number c such that P (c) = 0. Note that there will be some interval [a; b] such that P (a) and P (b) have oosite signs (and hence, we can choose k = 0), since the zoomed out behavior of P (x) will be the same as that of a nonzero constant times an odd ower of x. Examle 9: Here s how to nd a olynomial with integer coe cients that has 7 and 5 2i as zeros. To take care of 7, we ll use x 2 7 as a factor. To take care of 5 2i, begin by utting x = 5 2i, then isolate the ure imaginary art (this gives x 5 = 2i), then suare both sides (this gives (x 5) 2 = 4), then get a zero on one side (this gives (x 5) = 0). Therefore, a olynomial with integer coe cients that has 7 and 5 2i as zeros is x 2 7 h i (x 5) = x 4 10x + 22x x 20. there VII. Problems: Finding euations with seci ed comlex zeros Find olynomials with integer coe cients whose zeros include the given numbers. You may leave your answers in factored form. 1. A olynomial whose zeros include 5 and 2i. One answer: (x 5) (x ) A olynomial whose zeros include 5 + i and 2i. One answer: (x 5) (x ) A olynomial whose zeros include h 5 + i and. One answer: (x ) x x i A olynomial whose zeros include 6 5 and + 2 i. h One answer: x 2 6 i 2 45 (x ) [x (a + bi)] Q (x) = [x (a + bi)] [Q 1 (x) + iq 2 (x)]. Since the imaginary art of P (x) is zero, we have xq 2 (x) aq 2 (x) bq 1 (x) = 0. Solving for Q 1 gives Q 1 (x) = (x a)q 2(x). Plugging this exression b h i for Q 1 back into our reresentation of P (x) above gives P (x) = [x (a + bi)] (x a)q2 (x) + iq b 2 (x) = h i [x (a + bi)] x (a bi) Q 2 (x). Using this last exression, it is clear that P (a bi) = 0. b 10 Because olynomials of odd degree assume negative values for su ciently large negative inuts and ositive values for su ciently large ositive inuts.

6 VIII. Additional ractice roblems 1. [Refer to Examle 1 above] Use long division or synthetic division to exress 5x + (2 + 14i) x 2 + ( i) x + (15 5i) x i in the form a 0 + a 1 x + a 2 x 2 + R, where a 0, a 1, a 2, R 2 C. Show your work. 2. [Refer to Examle 2 above] Show the details involved with solving the four euations in Part III above using the method of Examle 2.. [Refer to Examles, 4, and 5 above] One way to factor a (b + c) 2 + b (c + a) 2 + c (a + b) 2 a 2 (b + c) b 2 (c + a) c 2 (a + b) is to exand this out and observe what you get. However, I want you to use the method illustrated in Examles, 4, and 5 to factor this exression. To get you started, note that the exression is zero if a = 0. Thus, a 0 = a is a factor. 4. [Refer to Examle 6 above] Show that is eual to 1 by showing this number is a solution to x + x x + x 4 to nd all solutions = 0 and then factoring 5. [Refer to Examle 6 above] Show that is eual to 2 by showing this number is a solution to x +6x x + 6x 20 to nd all solutions. 6. [Refer to Examle 6 above] Show that = 0 and then factoring is eual to by showing this number is a solution to x x 18 = 0 and then factoring x x 18 to nd all solutions This examle is taken from Claire Adler, A modern trick, American Mathematical Monthly 59 (1952), 28. This article is rerinted on of Tom M. Aostol, Gulbank D. Chakerian, Geraldine C. Darden, and John D. Ne (editors), Selected Paers on Precalculus, The Raymond W. Brink Selected Mathematical Paers #1, Mathematical Association of America, 1977 (MR 57 #280; Zbl ). 12 This examle is taken from Gary Slater, Solution to Problem 16.1, Mathematical Sectrum 16 # (198 84), The examle s 6 + r s6 r = 2 (a solution to x + 2x 12 = 0) is solved in two ways in Ira M. DeLong (editor), Solution to Problem #5, Problem Deartment column, School Science and Mathematics 7 #5 (May 1907), 41. The examle = 4 (a solution to x x 52 = 0) is solved in two ways in Solution to Problem #4190, Problem Deartment column, School Science and Mathematics 89 #4 (Aril 1989),

7 7. [Refer to Examles 7 and 8 above] Show the details involved with solving the ve euations in Part V above using the methods of Examles 7 and [Refer to Examle 9 above] Show the details involved with solving the four euations in Part VII above using the methods of Examle 9.

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