Download Polynomials and Algebraic Equations

STUDENT’S COMPANIONS IN MATH: NINETH
Polynomials and Algebraic Equations
A polynomial is a function of the form
p(x) = a0 xn + a1 xn−1 + · · · + an−1 x + an .
(1)
where a0 , a1 , a2 , . . . , an are some constants, called the coefficients of p. Using the
summation notation, we may write
p(x) =
n
X
ak xn−k .
k=0
In (1), when a0 6= 0, we say that the degree of p is n and we call a0 the leading
coefficient. When p is a nonzero constant, we say that the degree of p is zero. When
p is equal to the constant zero, it is customary assign ∞ to be the degree of p. We are
mainly interested in nonconstant polynomials and there is no need to worry about this
convention. By an algebraic equation we mean an equation of the form p(x) = 0. By a
root of the polynomial p, or a solution to the algebraic equation p(x) = 0, we mean
a (complex) number a such that p(a) = 0.
QUESTION 1. If 1 and 2 are roots of x3 − x2 − cx + d, what are c and d?
EXERCISE 2. Prove the following statement: if α and β are roots of x2 + ax + b and
α 6= β, then
−a = α + β,
b = αβ.
(2)
(Actually, the assumption α 6= β is unnecessary. We will mention a complete set of
identities giving the relation between roots and coefficients of a polynomial, traditionally
called Vieta’s formula; see (12) below.)
If the degree of p is two, we call p(x) = 0 a quadratic equation, if three, we call a cubic
equation, if four, we call a quatric equation, and if five, we call a quintic equation. Next we
discuss the general quadratic equation p(x) = 0, where p(x) = ax2 + bx + c with a 6= 0.
An important technique for dealing with a polynomial of degree two is called completing
square, which is based on the identity a2 + 2ab + b2 = (a + b)2 .
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EXAMPLE 3. We are asked to do the “completing square” for the following polynomials:
x2 + 6x + 7, 2x2 + 2x and 3x2 − 2x + 1. Here they are:
x2 + 6x + 7 = x2 + 2 . 3x + 32 − 32 + 7 = (x + 3)2 − 2,
2x2 + 2x = 2(x2 + x) = 2(x2 + 2(1/2)x + (1/2)2 − (1/2)2 ) = 2(x + 1/2)2 − 1/2,
3x2 − 2x + 1 = 3(x2 − 2(1/3)x + (1/3)2 − (1/3)2 ) + 1 = 3(x − 1/3)2 + 2/3.
As you can see, “completing square” is like a tailor’s job of fitting a given expression into
the form (x + a)2 , keeping but without worrying what is left over.
Now we return to the general polynomial of degree two: p(x) = ax2 + bx + c, with
a 6= 0. In the following discussion we assume that a, b, c are real numbers. We perform
“completing square” for this polynomial:
µ
¶
b
2
p(x) ≡ ax + bx + c = a x + x + c
a
Ã
½ ¾2 ½ ¾2 !
b
b
b
= a x2 + 2 x +
−
+c
2a
2a
2a
Ã
½ ¾2 !
b
b
b2
2
=a x +2 x+
−a 2 +c
2a
2a
4a
2
Or
µ
¶2
b
4ac − b2
p(x) = a x +
+
.
2a
4a
(3)
For convenience, let us write
p(x) = a(x − x0 )2 + m,
(4)
where x0 = −b/2a and m = (4ac − b2 )/4a. Notice that p(x0 ) = m and (x − x0 )2 is
always nonnegative. So, when a > 0, we have p(x) = a(x−x0 )2 +m ≥ 0+m = m = p(x0 ).
Thus, when a > 0, p has a unique (global) minimum m attained at x0 . Similarly,
when a < 0, we have p(x) = a(x − x0 )2 + m ≤ 0 + m = p(x0 ) and hence p has a
unique maximum m attained at x0 .
QUESTION 4. How do you use calculus to draw the same conclusion?
QUESTION 5. How do you use (3) or (4) to derive the following formula
x=
−b ±
√
b2 − 4ac
2a
2
(5)
for finding the roots of ax2 + bx + c = 0?
We return to the general theory of polynomials. You are assumed to know long division
for polynomials. See if you can do the following
EXERCISE 6. Use long division to divide f (x) = 2x4 − 7x3 + 14x + 4 by g(x) = x − 2.
Now we have to be more careful about numbers that we are allowed to use. To avoid
technicality, we restrict ourselves to the following three number systems: rationals, reals
and complex numbers. The standard notation for them is:
Q = the set of all rational numbers
R = the set of all real numbers
C = the set of all complex numbers.
In what follows, we use letter F (this is not a standard notation) to stand one of the
above: Q, R or C. We use this letter F here because it is a field according to some
technical definition. When a polynomial all coefficients a0 , a1 , . . . , an of a polynimial
p(x) = a0 xn + a1 xn−1 + · · · + an−1 x + an are in F, then we say that p is a polynomial
over F. Given polynomials f (x) and g(x) over F, we can divide f (x) by g(x), using
long division, to get a quotient q(x) and a remainder r(x), which are also polynomials over
F. The relation between f , g and q, r is given by the identity f (x) = g(x)q(x) + r(x).
Here, the degree of r(x) is strictly less than the degree of g(x), or r(x) is identically
zero. (Note: the degree of a nonzero constant polynomial is 0 but the degree of the zero
polynomial is defined to be infinity.) In case r(x) ≡ 0 so that we have f (x) = g(x)q(x),
we say that g is a factor of f or g divides f . When f has no factors over F other
than the trivial ones, that is, constants or constant multiples of f , we say that f is an
irreducible polynomial over F.
Notice that reducibility of a polynomial often depend on which field F we choose.
For example, x2 + 1 is a polynomial irreducible over Q or R. But it is reducible over
C because it has the following proper factorization: x2 + 1 = (x + i)(x − i).
QUESTION 7. Why is x2 − 2 is irreducible over Q but reducible over R?
Now we state the basic theorem concerning factorization of polynomials: A nonconstant polynomial can be written as a product of finitely many irreducible polynomials
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which are unique up to multiples of nonzero elements in F. The proof of this is similar
to the unique factorization of integers. A good book on abstract algebra should have a
theory (about something called PID, that is, principal ideal domain) covering the unique
factorization theorem for both polynomials and integers.
PROBLEM 8. Why is x4 + 1 irreducible over Q?
Recall that a number a is called a root of a polynomial if p(a) = 0. Given a
polynomial p(x) and a number a, we can divide p(x) by x − a to get p(x) =
(x − a)q(x) + r, where q is a polynomial with a degree smaller than that of p and the
remainder r is a number. Now suppose that a is a root of p(x): p(a) = 0. Letting
x = a in p(x) = (x − a)q(x) + r, we obtain 0 = r. Hence p(x) = (x − a)q(x). We have
shown that if a is a root of p(x), then x − a is a factor of p(x).
QUESTION 9. Why is the converse “if x − a is a factor of p(x), then a is a root of
p(x)” also true?
Next, suppose that b is another root: p(b) = 0 and b 6= a. Putting x = b in
p(x) = (x − a)q(x), we get 0 = (b − a)q(b). As b − a 6= 0, we have q(b) = 0. Thus b is
also a root of q(x) and hence x − b is a factor of q(x), say q(x) = (x − b)Q(x). Thus
p(x) = (x−a)(x−b)Q(x), telling us that (x−a)(x−b) is a factor of p(x). More generally,
if a1 , a2 , . . . , am are distinct roots of a polynomial p(x), then (x−a1 )(x−a2 ) · · · (x−am )
is a factor of p(x).
EXERCISE 10. Give a careful proof of the last assertion by inductiion on m.
One consequence of our discussion here is: If p(x) is a polynomial of degree n, then
p(x) cannot have more than n roots. To see this, suppose that p(x) has more than n
roots, say a1 , a2 , . . . , am with m > n. Then, according to what we have just learned,
f (x) ≡ (x − a1 )(x − a2 ) · · · (x − am ) is a factor of p(x). This cannot happen because the
degree of f (x) is m, which is greater than the degree of p(x), which is n. Indeed, the
degree of a factor of p cannot be greater than the degree of p.
Next we study the multiplicity of a root. Suppose that a is a root of a polynomial p,
that is, p(a) = 0. The above discussion tells that p(x) = (x − a)q(x) for some polynomial
q. Now we ask the question: is a also a root of q? The answer can be found by computing
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q(a), the value of q at a, to see if it is zero. If q(a) 6= 0, the answer is No and in that
case we say that a is a simple root of p. If q(a) = 0, the answer is Yes and in that
case we say that a is a multiple root of p.
EXERCISE 11. Prove that a is a simple root of p if and only if p(a) = 0 but p0 (a) 6= 0;
equivalently, a is a multiple root of a if and only if p(a) = 0 and p0 (a) = 0. (Of course,
p0 here stands for the derivative of p.)
Now we extract all factors of x − a from p. Say, there are m of them so that we can
write p(x) = (x − a)m g(x) for some polynomial g. Now g has no factor of x − a and
hence g(a) 6= 0. The positive integer m is called the multiplicity of the root a. When
m = 1, a is a simple root. When m > 1, a is a multiple root. When we count the number
of roots of a polynomial, we should count their multiplicities as well in order to get correct
answers to many questions.
PROBLEM 12. Prove by induction on m that the multiplicity of a root a of a polynomial
p is m if and only if p(k) (a) = 0 for 0 ≤ k < m and p(m) (a) 6= 0. (p(k) is the kth
derivative of p).
Concerning the existence of roots, we have the following very basic:
Fundamental Theorem of Algebra. If p is a polynomial of degree n ≥ 1 with
complex numbers as coefficients, then there is a complex number a (called a root of p)
such that p(a) = 0; in fact, counting multiplicities, p has exactly n roots.
Any proof of this basic theorem is beyond the level here, although there are over a hundred
of them; (I don’t know the recent record: I guess there are about two hundreds of proofs).
As we know, a real polynomial (that is, a polynomial with real coefficients) does not
necessarily have a real root. For a quadratic polynomial p(x) = ax2 + bx + c with real
coefficients a(6= 0), b, c, the recipe (5) in QUESTION 5 for its roots tells us that p does
not have real roots if its discriminant b2 − 4ac is negative and in that case its roots
are two complex numbers conjugate to each other. In general, we have the following fact
concerning the roots of a real polynomial:
Theorem. Suppose that p is a real polynomial of degree d ≥ 1. Then: (1) if the
degree s is an odd number, then p has at least one real root; (2) if ω is a nonreal root
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of p, then its complex conjugate ω is also a root of p; (3) p can be written as a product
of irreducible factors, either of the form rx + s or ax2 + bx + c, where r, s, a, b, c are
real numbers with r, a 6= 0 and b2 − 4ac < 0.
To prove this theorem, let p(x) = a0 xn + a1 xn−1 + · · · + an−1 x + an where a0 , a1 , . . . , an
are real numbers with a0 6= 0. If ω is a nonreal root of p, then p(ω) = 0 and hence
p(ω) = a0 ω n + a1 ω n−1 + · · · + an−1 ω + an
= a0 ω n + a1 ω n−1 + · · · + an−1 ω + an
(because the coefficients are real)
= p(ω) = 0 = 0
and hence ω is also a root of p. This proves (2). Now both x − ω and x − ω are
factors of p and they are different (otherwise ω = ω, contradicting the fact that ω is
nonreal). Thus (x − ω)(x − ω) is a factor of p. So we have p(x) = (x − ω)(x − ω)q(x)
for some polynomial q of degree n − 2. Now
(x − ω)(x − ω) = x2 + (ω + ω)x + ωω ≡ x2 + bx + c
with b = ω + ω, c = ωω real and b2 − 4ac = (ω + ω)2 − 4ωω = ω 2 + 2ωω + ω 2 − 4ωω =
(ω − ω)2 < 0.
QUESTION 13. Why are b and c real? Why is (ω − ω)2 negative?
Since p and (x−ω)(x−ω) ≡ x2 +bx+c are real polynomials and p(x) = (x2 +bx+c)q(x),
q is also a real polynomial.
QUESTION 14. Why?
We can ask if q has a nonreal root. If the answer is yes, we can get another conjugate
pair of nonreal roots and hence another factor of the form ax2 + bx + c with real a, b, c
such that b2 − 4ac < 0. Pulling out this factor from q (and hence from p), the degree
of f drops again by 2. Continue in this manner, until all nonreal roots are exhausted so
that only real roots of p remains. Now part (3) of the above theorem is clear. Part (1)
is a direct consequence of part (3).
From the above discussion, we see that irreducible polynomials over C are of the form
ax + b, where a 6= 0, and irreducible polynomials over R are of the form rx + s (r 6= 0)
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or ax2 + bx + c with b2 − 4ac < 0. What about irreducible polynomials over Q, the field
of rational numbers? This is hard, very hard. We can produce many of them by using
some advanced theorems such as Eisenstein’s criterion. Some of them such as cyclotomic
polynomials are very important. A complete description of irreducible polynomials over
rationals seems to be impossible.
EXAMPLE 15. We are asked to factorize the polynomial p(x) = x5 − 1 over the real
field R. Let ω = e2πi/5 . Then 1, ω, ω 2 , ω 3 ≡ ω 2 , z 4 = ω are roots of p. Thus
p(x) = (x − 1)(x − ω)(x − ω 2 )(x − ω 3 )(x − ω 4 )
= (x − 1)(x − ω)(x − ω)(x − ω 2 )(x − ω 2 )
(6)
= (x − 1)(x − (ω + ω)x + 1)(x − (ω 2 + ω 2 )x + 1),
in view of ωω = 1 and z 2 ω 2 = 1. Now please read EXAMPLE 17 of the SIXTH
√
COMPANION. From there we get ω + ω = 2 cos 2π/5 = ( 5 − 1)/2. So
Ã√
2
2
2
ω + ω = (ω + ω) − 2ωω =
5−1
2
!2
√
√
5−2 5+1
5+1
−2=
−2=−
.
4
2
Substituting the last expressions into (6), we have
Ã
!Ã
!
√
√
1
−
5
5
1
+
x5 − 1 = (x − 1) x2 +
x+1
x2 +
x+1 ,
2
2
which is the required factorization.
QUESTION 16. Are you sure the answer in the last example correct? Check the identity
Ã
!Ã
!
√
√
1− 5
1+ 5
x4 + x3 + x2 + x + 1 = x2 +
x+1
x2 +
x+1 .
2
2
EXERCISE 17. Factorize x4 + x2 + 1 over the real field R.
PROBLEM 18. Consider a polynomial of the form p(x) = xn + a1 xn−1 + · · · + an−1 x + an ,
where the coefficients a1 , a2 , . . . , an are integers. Prove that rational roots of p are
integers, that is, if r is a rational number such that p(r) = 0, then r is an integer. Use
√
this fact to deduce that, for positive integers m and n, n m is either an integer or an
√ √ √
irrational number. (In particular, 2,
3,
6 etc. are irrational.)
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In the rest we breifly describe other important aspects of polynomials. First: finding
roots by radicals. Formula (5) in QUESTION 5 is a closed form for roots of a general
quadratic equation in terms of a radical; (roughtly speaking, a radical is an algebraic expression involving roots, without trigonometric, exponential or other transcendental functions). For solving a general cubic equation x3 + ax2 + bx + c = 0, Cardano published
in 1545 a method described as follows. (This method was originally due to Tartaglia, to
whom Cardano promised earlier to keep it as a secret. Here we cannot give more detail of
this fascinating story.) First step: to eliminate the x2 term by a simple transformation
x = y + k, where k will be determined. Substituting x = y + k to x3 + ax2 + bx + c,
we get (y + k)3 + a(y + k)2 + · · · = y 3 + (3k + a)y 2 + · · · . Letting k = −a/3, the original
equation is converted into
y 3 + py + q = 0.
(7)
This step is more or less routine. It works for polynomial of any degree. The next step
is VERY SLICK: let y = u + v. (Here we replace one unknown y by two unknowns
u and v. But so far we only have one equation, namely (7). Therefore we have the
freedom to choose another equation!) Then (6) becomes (u + v)3 + p(u + v) + q = 0, or
u3 + 3u2 v + 3uv 2 + v 3 + p(u + v) + q = 0, that is,
u3 + v 3 + (3uv + p)(u + v) + q = 0.
(8)
The other equation we choose is 3uv + p = 0 so that (7) becomes u3 + v 3 + q = 0, or
u3 + v 3 = −q. From 3uv + p = 0 we obtain
uv = −p/3,
(9)
which gives u3 v 3 = −p3 /27. So, letting A = u3 and B = v 3 , we get
AB = −p3 /27
A + B = −q,
Solving this system, we get
q
A≡u =− +
2
3
r
q2
p3
+ ,
4
27
q
B≡v =− −
2
3
r
q2
p3
+ .
4
27
We get possible values of u and v:
u=
√
3
A, ω
√
3
A, ω 2
√
3
A
v=
8
√
3
B, ω
√
3
B, ω 2
√
3
B,
where ω = e2πi/3 . Due to the restrain (9), only three combinations of u and v are
possible for y = u + v as the solutions:
y=
√
3
A+
√
3
B, ω
√
3
A + ω2
√
3
B, ω 2
√
3
A+ω
√
3
B.
Quadric equations (algebraic equations of degree four) can be solved by the method of
reduction to cubic equations, which was discovered by Cardano’s student Ferrari. Great
efforts were made to solve quintic equations by radicals, but all of them were doomed to fail:
in early nineteen century Abel proved that this is impossible. In this case, merely using
radicals is not enough. To solve quintic equations one needs elliptic functions (such as the
Weierstrass P function ℘(z)) which are related to elliptic curves, a hot topic in current
mathematical research. The most powerful method of studying algebraic equations is the
Galois theory, which has tremendous impact on many other areas. Galois is a “romantic
revolutionary figure” in mathematics who died in a duel at the age of barely 21. [Another
“romantic figure” in math is Sofya Kowalevskaya, who is one of the two greatest women
in math and whose beauty is legendary. I’m just curious why Hollywood didn’t make a
movie about her.]
In EXERCISE 2 we have considered the relation between the roots and the coefficients
of a quadric. Consider a general polynomial p of degree n with leading coefficient 1:
p(x) = xn + a1 xn−1 + · · · + an−1 x + an .
(10)
Let x1 , x2 , . . . , xn be the roots of p. Then
p(x) = (x − x1 )(x − x2 ) · · · (x − xn ).
(11)
By multiplying out the right hand side of (11) and then comparing to (10), we have
a1 = −s1 , a2 = s2 , . . . an = (−1)n sn ,
where
s1 =x1 + x2 + · · · + xn ≡
X
k
xk
s2 =x1 x2 + x1 x3 + x2 x3 + · · · + xn−1 xn ≡
..
.
sn =x1 x2 · · · xn .
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X
j6=k
(12)
xj xk
(13)
The expressions s1 , s2 , . . . , sn , considered as functions of x1 , x2 , . . . xn , are called
elementary symmetric functions.
EXERCISE 19. Check (12) for n = 4.
A polynomial P = P (x1 , x2 , . . . , xn ) in n variables is called a symmetric polynomial if
it is unchanged after any permutation of its variables. For example, for n = 3, P (x1 , x2 , x3 )
is a polynomial if and only if it is invariant under six permutations of x1 , x2 , x3 :
P (x1 , x2 , x3 ) = P (x2 , x3 , x1 ) = P (x3 , x1 , x2 )
= P (x1 , x3 , x2 ) = P (x2 , x3 , x1 ) = P (x3 , x2 , x1 ).
For example, s1 , s2 , . . . , sn given by (13) are symmetric polynomials. Actually, they
generate all symmetric polynomials in the following sense:
Theorem. If P = P (x1 , x2 , . . . , xn ) is a symmtric polynomial, then there is a
polynomial Q in n variables such that P = Q(s1 , s2 , . . . , sn ), in other words, P can
be written as a polynomial in elementary symmetric functions. (This basic theorem about
symmetric polynomials, stated without proof here, is due to Newton.)
Given two polynomials f and g of degree m and n respectively, how do we
know if they have a common factor or they are relatively prime? You can use Euclidean’s
algorithm to find out, or compute a determinant of size m + n called the resultant (due
to Sylvester), denoted by R(f, g), to see if it vanishes. A polynomial p has a multiple
root if and only if p and its derivative p0 have a common factor. Hence R(p, p0 ) = 0
is the criterion for existence of multiple roots. The expression R(p, p0 ) is called the
discriminant of p. The discriminant of ax2 + bx + c turns out to be b2 − 4ac. This
should not surprise you! Resultant is also important for polynomials of several variables,
especially in a computation method called elimination.
We also mention that, for a real polynomial p, there is an important method due to
Sturm to determine how many real roots of p are lying in a given interval.
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