David Moon
Math 394
HW 1 Solutions
1.1 In class, we saw that given one root of a cubic polynomial, we can use the quadratic formula
to express the other two roots. The goal of this exercise is to generalize this principle.
(a) Carry out the above for the general (monic) cubic. In other words, given f (x) =
x3 + a2 x2 + a1 x + a0 , and β such that f (β) = 0, determine the other two roots of f (x)
in terms of β (and the ai ’s).
Use the general method in part (c) to get the quadratic polynomial
g(x) := x2 + (a2 − β)x + (a1 − β(a2 − β))x
and determine its roots.
(b) Now suppose you’re given the polynomial f (x) = x4 + a3 x3 + a2 x2 + a1 x + a0 , and
you happen to know that f (β) = 0. Write down a cubic polynomial whose roots are
precisely the other three roots of f (x).
Use the general method in part (c) to get the cubic polynomial
g(x) := x3 + (a3 − β)x2 + (a2 − β(a3 − β))x + (a1 − β(a2 − β(a3 − β))).
(c) Generalize part (b) to arbitrary monic polynomials of degree n.
P
Suppose polynomial f (x) := ni=0 ai xi (ai = 0) has roots α1 , α2 , . . . , αn . If we know
αn , we want to find the coefficients bi of the polynomial
g(x) = (x − α1 )(x − α2 ) . . . (x − αn−1 )
=
n−1
X
bi xi
(bn−1 = 1)
i=0
We can write
an−j = (−1)j
X
αi1 . . . αij ,
1≤i1 <···<ij ≤n
j
bn−j−1 = (−1)
X
αi1 . . . αij .
1≤i1 <···<ij ≤n−1
Observe that each bn−j−1 (the j th elementary symmetric function of α1 , . . . , αn−1 ) is
an−j (the j th elementary symmetric function of α1 , . . . , αn−1 , αn ) without any terms
that have αn . In each such term with αn , the j − 1 multiplicands other than αn
can vary freely in their indices—thus, factoring out αn from these terms gives us the
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(j − 1)th elementary symmetric function of α1 , . . . , αn−1 (that is, bn−(j−1)−1 ). More
succinctly, for each 1 ≤ j ≤ n − 1,
X
αi1 . . . αij
an−j = (−1)j
1≤i1 <···<ij ≤n


= (−1)j 
X
αi1 . . . αij +
= (−1)j 
αi1 . . . αij−1 αn 
1≤i1 <...ij−1 ≤n−1
1≤i1 <···<ij ≤n−1


X
X


αi1 . . . αij  + αn (−1)j 
X
αi1 . . . αij−1 
1≤i1 <...ij−1 ≤n−1
1≤i1 <···<ij ≤n−1
= bn−j−1 + αn bn−(j−1)−1
= bn−j−1 + αn bn−j .
Thus, we have the recursive definitions
bn−j−1 = an−j − αn bn−j
for each 1 ≤ j ≤ n − 1, which are well-defined because bn = 1.
(Note that this formulation recurses downward from bn to b1 . Some of you essentially
recursed upward starting from b0 = −a0 /αn , but this requires division by αn and thus
extra work in the case αn = 0.)
1.2 Given a, b, c ∈ C. Prove that there exist α, β, γ ∈ C such that


a = α + β + γ
b = α + βω + γω 2


c = α + βω 2 + γω
where ω = e2πi/3 . Are α, β, γ uniquely determined by the system above?
We have

   
1 1
1
α
a
1 ω ω 2   β  =  b  .
1 ω2 ω
γ
c
We can compute a nonzero determinant for the matrix above, which implies the existence
and uniqueness of the solution.
1.3 In class we discovered how to find the roots of the polynomial x3 − 6x + 2.
(a) Generalize our method and explicitly determine the solutions to the equation x3 −3px+
2q = 0, where p, q are complex numbers. [Of course you could look up the formula,
but this is not instructive. Instead, I ask you to derive it using the same approach we
used in class.]
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Let a, b, c be the roots of polynomial x3 − 3px + 2q. Then
a + b + c = 0,
(1)
ab + bc + ca = −3p,
(2)
abc = −2q.
(3)
By Exercise 1.2, there exist unique α, β, γ ∈ C such that
a = α + β + γ,
(4)
2
b = α + βω + γω ,
(5)
c = α + βω 2 + γω.
(6)
These and (1) give us
0=a+b+c
= 3α + (1 + ω + ω 2 )β + (1 + ω + ω 2 )γ
= 3α,
so α = 0. Rewriting (4) (5) (6), we have
a = β + γ,
(7)
2
b = βω + γω ,
(8)
2
(9)
c = βω + γω.
These and (2) give us
−3p = ab + bc + ca
= (β 2 ω + βγ(ω + ω 2 ) + γ 2 ω 2 ) + (β 2 + βγ(ω + ω 2 ) + γ 2 ) + (β 2 ω 2 + βγ(ω + ω 2 ) + γ 2 ω)
= β 2 (1 + ω + ω 2 ) + γ 2 (1 + ω + ω 2 ) + 3βγ(ω + ω 2 )
= −3βγ,
and these and (3) give us
−2q = abc
= (β + γ)(βω + γω 2 )(βω 2 + γω)
..
.
= β3 + γ3.
Constructing the auxiliary quadratic
(t − β 3 )(t − γ 3 ) = t2 − (β 3 + γ 3 )t + β 3 γ 3 = t2 + 2qt + p3
gives us
3
3
β ,γ =
−2q ±
p
p
4q 2 − 4p3
= −q ± q 2 − p3 .
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(b) What are the roots of x3 − 3x? Now find them by applying the above formula to
determine the roots of x3 − 3x. This type of computation was the original (historical)
impetus for the introduction of i.
This is a straightforward computation.
(c) Explicitly determine one solution of the equation x3 − 6x2 + 21x − 22 = 0.
We can rewrite the cubic in its depressed form (x − 2)3 + 9(x − 2) + 4, at which point
we can apply our work above to compute a solution.
1.4 Use Euler’s formula eiθ = cos θ + i sin θ to discover the triple angle formulas (i.e. formulas
for sin 3θ and cos 3θ in terms of sin θ and cos θ).
We have
cos 3θ + i sin 3θ = e3iθ
= (eiθ )3
= (cos θ + i sin θ)3
= cos3 θ + 3i cos2 θ sin θ + 3i2 cos θ sin2 θ + i3 sin3 θ
= (cos3 θ − 3 cos sin2 θ) + i(3 cos2 θ sin θ − sin3 θ).
Thus,
cos 3θ = cos3 θ − 3 cos sin2 θ,
sin 3θ = 3 cos2 θ sin θ − sin3 θ.
1.5 In this exercise you will explore some properties of complex roots of unity (i.e. solutions to
equations of the form xn = 1).
(a) Recall that ω = e2πi/3 is a cube root of unity, since it is a solution of the equation
x3 − 1 = 0. Is there a polynomial f (x) ∈ Z[x] of degree smaller than 3 which also has
ω as a root?
Yes. Recall our favorite identity ω 2 +ω +1 = 0. In other words, ω is a root of x2 +x+1.
(b) Find the monic polynomial in Z[x] of smallest degree which has e2πi/5 as a root.
We know α := e2πi/5 is a fifth root of unity, i.e., α is a root of x5 − 1. Since
x5 − 1 = (x − 1)(x4 + x3 + x2 + x + 1)
(and Z—and therefore Z[x]—is a UFD), we have α is a root of g(x) := x4 + x3 + x2 +
x + 1.
To show g(x) has minimal degree, one approach is to show that α cannot satisfy a
polynomial of lower degree. A less computation-intensive, more general approach is to
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observe that g(x) is irreducible in Z[x] by Eisenstein’s criterion, and then show that
any irreducible polynomial g(x) with a root α has minimal degree over all polynomials
having α as a root.
First consider an arbitrary polynomial h(x) ∈ Z[x] of minimal degree such that h(α) =
0. Taking any other polynomial f (x) ∈ Z[x] such that f (α) = 0, we can use the
Euclidean division algorithm to write
f (x) = h(x)q(x) + r(x)
for some polynomials q(x), r(x) ∈ Z[x] such that deg r(x) < deg h(x). Plugging in α,
we have that r(α) = 0, which contradicts the minimality of the degree of h(x) unless
r(x) = 0. Thus, we have that h(x) divides any other polynomial in Z[x] that has α
as a root. Since our polynomial g(x) is irreducible, we conclude that g(x) must also
have minimal degree.
(c) Fix a prime p. Expand and simplify (with proof !) the product
p−1
Y
(x − e2πik/p ).
k=1
We know e2πik/p for all 1 ≤ k ≤ p − 1 are p − 1 of the pth roots of unity. Then
(x − 1)
p−1
Y
(x − e2πik/p ) = xp − 1 = (x − 1)(xp−1 + xp−2 + · · · + x + 1),
k=1
which gives us our desired polynomial xp−1 + xp−2 + · · · + x + 1 since C[x] is a UFD.
1.6 We guessed that the three roots of a cubic have the form
α + β + γ,
α + βω + γω 2 ,
α + βω 2 + γω
What would you guess to be the form of the four roots of a quartic?
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