Download Complex Conjugation and Polynomial Factorization I. The Remainder

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
x 4i
3i)
=
14 + 2ix +
4 59i
x 4i
Also, you should be able to verify by direct substitution that P (4i) = 4
Proof:
59i.
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
(x + y + z) (xy + yz + xz)
0) is a factor of
(x + y) (y + z) (x + z)
since putting x = 0 in this expression gives zero. [Suppose y = 5 and z = 2. Then the expression becomes (x + 3) (5x
10
2x)
x is a factor of (x + y + z) (xy + yz + xz)
(x + 5) (3) (x
2), and this equals zero when x = 0. Thus,
(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.]
1
Keep in mind that every real number is also a complex number. Real numbers are just complex numbers
that have the form a + 0 i.
2
Do you know why R will be a constant?
Example 4:
We can see that m + n is a factor of
m4
4m3 n + 2m2 n2 + 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
5m2 n +
2
3
7mn
2n .
Example 5:
Suppose we want to factor
(x + y + z)3
x3 + y 3 + z 3
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 + y 3 + z 3
= 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.
p
p
p
p
3
3
3 21 + 8
3 21
8 = 1,
Example 6: By using your calculator it would appear that
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
2.
x3
3.
x4 10x3 +70x2 +40x 296 = 0, given that x =
4.
20x
16 = 0, given that x = 4 is a solution. Answer: x = 4;
6x2 = 90x + 148, given that x =
2 is a solution. Answer: x =
p
5.
3
2; 4
p
3 10 .
2 are solutions. Answer: x =
x5 x4 +x = 10x3 10x2 +1, given that x = 1 is a solution. Answer: x = 1;
p
5
2; 5
This 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.
4
Because dividing (x + y + z)3
x3 + y 3 + z 3 by (x + y) (y + z) (x + z) must give an algebraic expression of zero degree, and only constants have zero degree.
p
p
p
p
5
Compare this to, for example, to the task of showing that
3 +2 2
3
2 2 is equal to 2,
where expanding and simplifying the square of this expression will work.
6
Verifying
that this number
is a solution to x3 + 15x 16 = 0 is not as tedious as you might think. Let
p
p
p
p
3
3
x =
3 21 + 8
3 21
8 . Using the identity (a b)3 = a3 3a2 b + 3ab2 b3 and short–cuts
p
p
p
p
p
p
2
such as 3 21 + 8
3 21
8 = 3 21 + 8 3 21
8
3 21 + 8 = 125 3 21 + 8 , it
q
q
p
p
doesn’t take much to show that x3 = 16
3 3 125 3 21 + 8 + 3 3 125 3 21
8 . But this is just
p
p
p
p
3
3
3
16
15 3 21 + 8 + 15 3 21
8 , which is clearly equal to 16 15x. Thus, x = 16 15x, and so
x3 + 15x 16 = 0.
3
7i.
p
2 6 .
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:
p
p
p
p
x
2 x
3 is a polynomial with 2 and 3 as zeros. We can …nd a polyp
p
nomial with integer coe¢ cients having 2 and 3 as zeros if we pair each of these
factors with their conjugates
way. Note that what I’m doing amounts
p
p in the following
to choosing P (x) = x + 2 x + 3 .
x
p
2
x+
p
2
x
p
3
x+
p
3
=
x2
2
x2
3
=
x4 5x2 +6
A more
way to obtain this “nice” polynomial is to …nd “nice” polynomials
p systematic
p
for 2 and p3 separately and then multiply these two “nice” polynomials. Begin by
putting x = 2 , then square both sides, and then subtract 2 from both sides to p
get
2
a 0 on one side. The other side will be a polynomial
(speci…cally,
x
2)
that
has
2
p
as a zero. If you do the same thing for x = 3 , you’ll get x2 3.
p
p
Example 8: Here’s how to …nd a polynomial with integer coe¢ cients that has 2 + 3 as a zero.
p
p
p
p
Let x = 2 + 3 and square both sides. This gives x2 = 5 + 2 6 . Now isolate 2 6
p 2
2
and square again: x2 5 = 2 6 . The polynomial we want will be the polynomial 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
p
p
verify by hand calculation, each of the four sign choices in
2
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 ),
p
p
you’ll get x =
5
2 6 . Note this is consistent with the third sentence in this example, where
p
p
p
p
p
p
p
I wrote x2 = 5 + 2 6 . Thus,
5 + 2 6 must equal
2 + 3,
5
2 6
must equal
p
p
2 + 3 , etc. In older college algebra texts (atp
least 50 years old) you can often …nd a discussion
p
of techniques for rewriting expressions of the form A + B as a sum or di¤erence of square roots.
p
p
p
p
This was useful because, for instance, 2 + 3 tends to be easier to work with than 5 + 2 6 ,
p
p
p
p
plus it’s a lot easier to obtain a numerical approximation for 2 + 3 than it is for 5 + 2 6 .
p
p
p
p
In the case of 2 + 3 , you simply looked up the values of 2 and 3 in a table and added the
p
p
p
p
approximate values by hand. A more exotic example is 59 90
14 7 + 4 4555 + 1721 7
p
p
being equal to 145 26 + 2 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
2)2
One answer: (x
2.
3.
2)2
4.
5
2
2)
5
2
(x + 4)
5 and
p
5,
One answer: x + 9x
2
2
2
p
3
2 +
One answer:
2
x
6
2
i2
45
4, and
p
2.
p
3.
2
27 x + 1 .
q
6
A polynomial whose zeros include
h
4.
2 .
x
A polynomial whose zeros include
3
5.
p
(x + 4).
A polynomial whose zeros include 2
One answer: (x
5.
5.
A polynomial whose zeros include 2
One answer: (x
p
162.
3
p
5+
p
2 .
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
p
p
Hint for V(4): Beginning with x = 3 2 + 3 , isolate the
p simplify
p cube root, then cube both sides, then
all numerical square roots so that they will be multiples of 3 , then isolate all terms involving 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:
s
r
q
q
p
p
p
p
1
sin
=
34
2 17
2 34
2 17
4 17 + 3 17
170 + 38 17
17
8
7
8
More 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.
9
Here’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:
p
Here’s how to …nd a polynomial
with
integer
coe¢
cients
that
has
7 and 5 2i as
p
zeros. To take care of 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
p
gives (x 5)2 + 4 = 0). Therefore,
h a polynomial
i with integer coe¢ cients that has 7
and 5 2i as zeros is x2 7
(x 5)2 + 4 = 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
One answer: (x
2.
5)
5)2 + 1
2i.
3)2 + 4 .
(x
p
A polynomial whoseh zeros include i 5 + i and 3.
2
One answer: (x
4.
2i.
3)2 + 4 .
A polynomial whose zeros include 5 + i and 3
One answer: (x
3.
(x
x2 + 6
3)
20x2 .
A polynomial
whose zerosiinclude
h
One answer:
2
x
6
2
45
(x
p
6
p
3 5
and 3 +
p
2 i.
2
3) + 2 .
[x
(a + bi)] Q (x) = [x
(a + bi)] [Q1 (x) + iQ2 (x)]. Since the imaginary part of P (x) is zero, we have
2 (x)
xQ2 (x)
aQ2 (x)
bQ1 (x) = 0. Solving for Q1 gives Q1 (x) = (x a)Q
b
h . Plugging this expression
i
2 (x)
for Q1 back into our representation of P (x) above gives P (x) = [x
(a + bi)] (x a)Q
+ iQ2 (x)
b
h
i
[x
(a + bi)] x (ab bi) Q2 (x). Using this last expression, it is clear that P (a bi) = 0.
10
=
Because 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
x
3i
3i) x + (15
5i)
in the form a0 + a1 x + a2 x2 + 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 factor
a (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
q
p
3
2 + 5
q
2 +
[Refer to Example 6 above] Show that
q
p
3
10 + 108
q
10 +
3
p
5
is equal to 1 by showing this number is a solution to x3 + 3x
x3 + 3x 4 to …nd all solutions.11
5.
3
p
4 = 0 and then factoring
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 that
q
p
3
9 + 4 5
+
q
3
9
is equal to 3 by showing this number is a solution to x3
x3 3x 18 to …nd all solutions.12
p
4 5
3x 18 = 0 and then factoring
11
This 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).
12
This example is taken from Gary Slater, Solution to Problem 16.1, Mathematical Spectrum 16 #3 (1983–
84), 96–97. The example
s
s
r
r
3
3
980
980
6 +
+
6
= 2
27
27
(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
q
q
p
p
3
3
26 + 675
+
26
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.
7.
[Refer to Examples 7 and 8 above] Show the details involved with solving the …ve equations
in Part V above using the methods of Examples 7 and 8.
8.
[Refer to Example 9 above] Show the details involved with solving the four equations in
Part VII above using the methods of Example 9.