Complex Conjugation and Polynomial Factorization
Dave L. Renfro Summer 2004
Central Michigan University
I. The Remainder Theorem
Let P (x) be a polynomial with complex coe¢ cients1 and r be a complex number. The remainder when P (x) is divided by x r is P (r).
Example 1: If P (x) = 2ix2 6x + (4 3i) and r = 4i, then by long division or by synthetic division you should be able to verify that
2ix2 6x + (4 3i)
x 4i = 14 + 2ix + 4 59i
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 “quotient + remainder” form, we have P (x) = (x r) Q (x) + R, where Q (x) is the quotient and R (a constant2) is the remainder. Now plug x = r into both sides. This gives P (r) = R, which is what we were to prove.
II. The Factor Theorem
Let P (x) be a polynomial with complex coe¢ cients suppose r is a complex 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 special case of the Remainder Theorem.
Example 2: If P (x) = x3 12x2+ 94x 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) x2 10x + 74 . Note that armed with this factorization, you can …nd the other two zeros of this polynomial: x = 5 7i.
Example 3: 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 putting x = 0 in this expression gives zero. [Suppose y = 5 and z = 2. Then the ex- pression becomes (x + 3) (5x 10 2x) (x + 5) (3) (x 2), and this equals 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 expression for any two other choices of y and z, since no matter what y and z are, the expression algebraically reduces to zero when x = 0. Hence, x is a factor of this expression no matter what y and z are.]
1Keep in mind that every real number is also a complex number. Real numbers are just complex numbers that have the form a + 0 i.
2Do you know why R will be a constant?
Example 4: We can see that m + n is a factor of
m4 4m3n + 2m2n2 + 5mn3 2n4
since plugging n in for m in this expression gives zero. If you use long division or synthetic division, you’ll …nd the quotient after dividing by m + n is m3 5m2n + 7mn2 2n3.
Example 5: Suppose we want to factor
(x + y + z)3 x3+ y3+ z3
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.3 Since this product has the same degree as the expression we’re trying to factor, it follows that
(x + y + z)3 x3+ y3+ z3 = k (x + y) (y + z) (x + z)
for some constant k.4 To …nd k, we simply plug in three numbers and solve for k.
For example, x = 0, y = 1, and z = 2 gives us (0 + 1 + 2)3 03+ 13+ 23 = k (0 + 1) (1 + 2) (0 + 2). Therefore, 18 = 6k, and hence k = 3.
Example 6: By using your calculator it would appear that p3 3p
21 + 8 p3 3p
21 8 = 1,
but how could we prove this?5 One way is to verify by hand that this number is a solution to x3+ 15x 16 = 0 and then use the fact that x = 1 is the only real solution of x3+ 15x 16 = 0, since x3+ 15x 16 = (x 1) x2+ x + 16 .6 I challenge you to
…nd another way to prove this number is 1!
III. Using the Factor Theorem to solve equations
1.
x3+ 2x2 20x 16 = 0, given that x = 4 is a solution.Answer: x = 4; 3 p 5.2.
x3 6x2 = 90x + 148, given that x = 2 is a solution.Answer: x = 2; 4 3p 10.3.
x4 10x3+70x2+40x 296 = 0, given that x = 2 are solutions.Answer: x = 2; 5 7i.4.
x5 x4+x = 10x3 10x2+1, given that x = 1 is a solution.Answer: x = 1; p 5 2p6 .
3This is because no two of x + y, y + z, and x + z share any factors. Note that (x y)2 and (x y)3 are factors of (x y)4, but their product isn’t.
4Because dividing (x + y + z)3 x3+ y3+ z3 by (x + y) (y + z) (x + z) must give an algebraic expres- sion of zero degree, and only constants have zero degree.
5Compare this to, for example, to the task of showing that p 3 + 2p
2 p
3 2p
2 is equal to 2, where expanding and simplifying the square of this expression will work.
6Verifying that this number is a solution to x3+ 15x 16 = 0is not as tedious as you might think. Let x = p3
3p
21 + 8 p3 3p
21 8. Using the identity (a b)3 = a3 3a2b + 3ab2 b3 and short–cuts such as 3p
21 + 8 2 3p
21 8 = 3p
21 + 8 3p
21 8 3p
21 + 8 = 125 3p
21 + 8 , it doesn’t take much to show that x3 = 16 33
q 125 3p
21 + 8 + 33 q
125 3p
21 8 . But this is just 16 15p3
3p
21 + 8 + 15p3 3p
21 8, which is clearly equal to 16 15x. Thus, x3= 16 15x, and so x3+ 15x 16 = 0.
IV. Finding a polynomial with speci…ed real zeros
If a polynomial has zeros r1 and r2, then (x r1) (x r2) is a polynomial that has r1 and r2 as zeros. In fact, if P (x) is any polynomial, then P (x) (x r1) (x r2) will also be a polynomial that has r1 and r2 as zeros, although the presence of P (x) will likely generate additional zeros.
Example 7: x p
2 x p
3 is a polynomial with p
2 and p
3 as zeros. We can …nd a poly- nomial with integer coe¢ cients having p
2 and p
3 as zeros if we pair each of these factors with their conjugates in the following way. Note that what I’m doing amounts to choosing P (x) = x +p
2 x +p 3 .
x p
2 x +p
2 x p
3 x +p
3 = x2 2 x2 3 = x4 5x2+6 A more systematic way to obtain this “nice” polynomial is to …nd “nice” polynomials forp
2 and p
3 separately and then multiply these two “nice” polynomials. Begin by putting x =p
2 , then square both sides, and then subtract 2 from both sides to get a 0 on one side. The other side will be a polynomial (speci…cally, x2 2) that hasp
2 as a zero. If you do the same thing for x =p
3 , you’ll get x2 3.
Example 8: Here’s how to …nd a polynomial with integer coe¢ cients that hasp 2 +p
3 as a zero.
Let x = p 2 +p
3 and square both sides. This gives x2 = 5 + 2p
6 . Now isolate 2p 6 and square again: x2 5 2 = 2p
6 2. The polynomial we want will be the polyno- mial that shows up when we get a zero on one side of the equation, which is x4 10x2+1.
Since this is a 4’th degree polynomial, there should be four zeros, counting multiplicity. As you can verify by hand calculation, each of the four sign choices in p
2 p
3 give a zero of this polynomial.
On the other hand, if you solve this by using the quadratic formula (the equation is quadratic in x2), you’ll get x = p
5 2p
6 . Note this is consistent with the third sentence in this example, where I wrote x2 = 5 + 2p
6. Thus, p 5 + 2p
6 must equal p 2 +p
3, p 5 2p
6 must equal p2 +p
3, etc. In older college algebra texts (at least 50 years old) you can often …nd a discussion of techniques for rewriting expressions of the formp
A +p
B as a sum or di¤erence of square roots.
This was useful because, for instance, p 2 +p
3 tends to be easier to work with thanp 5 + 2p
6 , plus it’s a lot easier to obtain a numerical approximation for p
2 +p
3 than it is forp 5 + 2p
6 . In the case ofp
2 +p
3, you simply looked up the values ofp
2 and p
3 in a table and added the approximate values by hand. A more exotic example is 59p
90 14p
7 + 4p
4555 + 1721p 7 being equal to 145p
26 + 2p 7 .
V. Problems: Finding equations with speci…ed real zeros
Find polynomials with integer coe¢ cients whose zeros include the given numbers. You may leave your answers in factored form.7
1.
A polynomial whose zeros include 2 p 5 .One answer: (x 2)2 5.
2.
A polynomial whose zeros include 2 p5 and 4.
One answer: (x 2)2 5 (x + 4).
3.
A polynomial whose zeros include 2 p5 , 4, andp 2 .
One answer: (x 2)2 5 (x + 4) x2 2 .
4.
A polynomial whose zeros include p3 2 +p3 .
One answer: x3+ 9x 2 2 27 x2+ 1 2.
5.
A polynomial whose zeros include q6 3p 5 + p
2 .
One answer: h
x2 6 2 45i2
162.
VI. Complex zeros of polynomials with real coe¢ cients
If a polynomial has real coe¢ cients, then its zeros occur in complex conjugate pairs.8 I went through the details of this in class. The basic idea is that complex conjugation distributes through addition and multiplication (i.e. (z1+ z2) = z1 + z2 and (z1 z2) = z1 z2), and from this you can show that P (z) = P (z) if P is a polynomial with real coe¢ cients.
Therefore, if P (c) = 0 (i.e. c is a zero of P ) then, by taking the complex conjugate of both sides, we get P (c) = 0, which implies P (c) = 0 (i.e. c is a zero).9
7Hint for V(4): Beginning with x = p3 2 +p
3, isolate the cube root, then cube both sides, then simplify all numerical square roots so that they will be multiples ofp
3, then isolate all terms involving p
3 on one side of the equation, then square both sides, then get a zero on one side.
Since I have a little extra room on this page, here’s a numerical marvel for your amusement:
sin 17 = 1
8 s
34 2p
17 2
q
34 2p
17 4
r
17 + 3p 17
q
170 + 38p 17
8More generally, we have the following result. Let P be a non–constant polynomial with complex coe¢ - cients. Then the zeros of P occur in complex conjugate pairs if and only if there exists a line L in the complex plane such that 0 2 L and all the coe¢ cients of P belong to L. This is Monthly Problem E 1133 (proposed by Murray R. Spiegel), whose solution (by F. D. Parker) is given in American Mathematical Monthly 62 #4 (April 1955), 256–257: If the coe¢ cients lie on a common line, = 1, through the origin, then writing the coe¢ cients in polar form produces a factor ei 1. When this factor is divided out, the coe¢ cients are real and the imaginary [non–real] roots are in conjugate pairs. If the imaginary roots occur in conjugate pairs, then an equation with real coe¢ cients can be formed with those roots. Any other equation with the same roots must di¤ er by only a multiplicative constant. Any such constant will place the new coe¢ cients on a common line through the origin.
9Here’s an interesting alternative method given by D. Trifan in Complex roots of an integral rational equation, Amer. Math. Monthly 61 (1954), 640–641. If P (a + bi) = 0, then we can factor P (x) as
Application: If P (x) is an odd–degree polynomial with real coe¢ cients, then P (x) has at least one real zero.
A “calculus way” to prove this same result is to observe that polynomials are continuous on every interval, and hence the Intermediate Value Theorem holds on every interval, which implies1 0 there 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 opposite 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 power of x.
Example 9: Here’s how to …nd a polynomial with integer coe¢ cients that has p
7 and 5 2i as zeros. To take care of p
7 , we’ll use x2 7 as a factor. To take care of 5 2i, begin by putting x = 5 2i, then isolate the pure imaginary part (this gives x 5 = 2i), then square both sides (this gives (x 5)2 = 4), then get a zero on one side (this gives (x 5)2+ 4 = 0). Therefore, a polynomial with integer coe¢ cients that hasp
7 and 5 2i as zeros is x2 7 h
(x 5)2+ 4i
= x4 10x3+ 22x2+ 70x 203.
VII. Problems: Finding equations with speci…ed complex zeros
Find polynomials with integer coe¢ cients whose zeros include the given numbers. You may leave your answers in factored form.
1.
A polynomial whose zeros include 5 and 3 2i.One answer: (x 5) (x 3)2 + 4 .
2.
A polynomial whose zeros include 5 + i and 3 2i.One answer: (x 5)2 + 1 (x 3)2 + 4 .
3.
A polynomial whose zeros include p5 + i and 3.
One answer: (x 3) h
x2+ 6 2 20x2i .
4.
A polynomial whose zeros include p6 3p
5 and 3 + p 2 i.
One answer: h
x2 6 2 45i
(x 3)2 + 2 .
[x (a + bi)] Q (x) = [x (a + bi)] [Q1(x) + iQ2(x)]. Since the imaginary part of P (x) is zero, we have xQ2(x) aQ2(x) bQ1(x) = 0. Solving for Q1 gives Q1(x) = (x a)Qb 2(x). Plugging this expression for Q1 back into our representation of P (x) above gives P (x) = [x (a + bi)]h(x a)Q
2(x)
b + iQ2(x)i
= [x (a + bi)]hx (a bi)
b
iQ2(x). Using this last expression, it is clear that P (a bi) = 0.
1 0Because polynomials of odd degree assume negative values for su¢ ciently large negative inputs and positive values for su¢ ciently large positive inputs.
VIII. Additional practice problems
1.
[Refer to Example 1 above] Use long division or synthetic division to express 5x3 + (2 + 14i) x2 + ( 3 3i) x + (15 5i)x 3i
in the form a0 + a1x + a2x2 + R, where a0, a1, a2, R 2 C. Show your work.
2.
[Refer to Example 2 above] Show the details involved with solving the four equations in Part III above using the method of Example 2.3.
[Refer to Examples 3, 4, and 5 above] One way to factora (b + c)2 + b (c + a)2 + c (a + b)2 a2(b + c) b2(c + a) c2(a + b) is to expand this out and observe what you get. However, I want you to use the method illustrated in Examples 3, 4, and 5 to factor this expression. To get you started, note that the expression is zero if a = 0. Thus, a 0 = a is a factor.
4.
[Refer to Example 6 above] Show that q32 + p
5 3
q
2 + p 5
is equal to 1 by showing this number is a solution to x3+ 3x 4 = 0 and then factoring x3+ 3x 4 to …nd all solutions.11
5.
[Refer to Example 6 above] Show that3
q
10 + p
108 3
q
10 + p 108
is equal to 2 by showing this number is a solution to x3+6x 20 = 0 and then factoring x3+ 6x 20 to …nd all solutions.
6.
[Refer to Example 6 above] Show that3
q
9 + 4p
5 + 3
q
9 4p 5
is equal to 3 by showing this number is a solution to x3 3x 18 = 0 and then factoring x3 3x 18 to …nd all solutions.12
1 1This example is taken from Claire Adler, A modern trick, American Mathematical Monthly 59 (1952), 328. This article is reprinted on pp. 276–277 of Tom M. Apostol, Gulbank D. Chakerian, Geraldine C.
Darden, and John D. Ne¤ (editors), Selected Papers on Precalculus, The Raymond W. Brink Selected Mathematical Papers #1, Mathematical Association of America, 1977 (MR 57 #2803; Zbl 357.00002).
1 2This example is taken from Gary Slater, Solution to Problem 16.1, Mathematical Spectrum 16 #3 (1983–
84), 96–97. The example
3
s 6 +
r 980
27 + 3
s 6
r980
27 = 2
(a solution to x3+ 2x 12 = 0) is solved in two ways in Ira M. DeLong (editor), Solution to Problem #53, Problem Department column, School Science and Mathematics 7 #5 (May 1907), 413. The example
q3
26 + p
675 + 3
q
26 p
675 = 4
(a solution to x3 3x 52 = 0) is solved in two ways in Solution to Problem #4190, Problem Department column, School Science and Mathematics 89 #4 (April 1989), 357–358.
