Practice 107
709 Example
The polynomial px has integral coefficients and px = 7 for four different values of x. Shew that px never equals 14.
Solution: The polynomial gx = px − 7 vanishes at the 4 different integer values a , b, c, d. In virtue of the Factor Theorem,
gx = x − ax − bx − cx − dqx ,
where qx is a polynomial with integral coefficients. Suppose that pt = 14 for some integer t. Then gt = pt − 7 = 14 − 7 = 7
. It follows that 7 = gt = t − at − bt − ct − dqt
, that is, we have factorised 7 as the product of at least 4 different factors, which is impossible since 7 can be factorised as
7−11, the product of at most 3 distinct integral factors. From this contradiction we deduce that such an integer t does not exist.
Practice
710 Problem If px is a polynomial of degree n such that
pk = 1 k, k = 1, 2, . . . ,n + 1, find pn + 2.
711 Problem The polynomial px satisfies p−x = −px
. When px is divided by x − 3 the remainder is 6. Find the remainder when px is divided by x
2
− 9.
6.4 Viète’s Formulae
Let us consider first the following example.
712 Example Expand the product
x + 1x − 2x + 4x − 5x + 6 .
Solution: The product is a polynomial of degree 5. To obtain the coefficient of x
5
we take an x from each of the five binomials. Therefore, the coefficient of x
5
is 1. To form the x
4
term, we take an x from 4 of the binomials and a constant from the remaining binomial. Thus the coefficient of x
4
is 1 − 2 + 4 − 5 + 6 = 4
. To form the coefficient of x
3
we take three x from 3 of the binomials and two constants from the remaining binomials. Thus the coefficient of x
3
is 1−2 + 14 + 1−5 + 16 + −24 + −2−5 + −26
+4−5 + 46 + −56 = −33 .
Similarly, the coefficient of x
2
is 1−24 + 1−2−5 + 1−26 + 14−5 + 146 + −24−5
+−246 + 4−56 = −134 and the coefficient of x is
1−24−5 + 1−246 + 1−2−56 + 14−56 + −24−56 = 172 .
Finally, the constant term is 1−24−56 = 240 . The product sought is thus
x
5
+ 4x
4
− 33x
3
− 134x
2
+ 172x + 240 .
From the preceding example, we see that each summand of the expanded product has “weight” 5, because of the five given binomials we either take the x or take the constant.
108 Chapter 6
If a 6= 0 and
a x
n
+ a
1
x
n−1
+ a
2
x
n−2
+ ··· + a
n−1
x + a
n
is a polynomial with roots
α
1
,
α
2
, . . . ,
α
n
then we may write a
x
n
+ a
1
x
n−1
+ a
2
x
n−2
+ ··· + a
n−1
x + a
n
= a x −
α
1
x −
α
2
x −
α
3
··· x −
α
n−1
x −
α
n
. From this we deduce the Viète Formulæ:
− a
1
a =
n
X
k=1
α
k
, a
2
a =
X
1 ≤ jk≤n
α
j
α
k
, −
a
3
a =
X
1 ≤ jkl≤n
α
j
α
k
α
l
, a
4
a =
X
1 ≤ jkls≤n
α
j
α
k
α
l
α
s
, ..........
.......... ...........
−1
n
a
n
a =
α
1
α
2
···
α
n
.
713 Example
Find the sum of the roots, the sum of the roots taken two at a time, the sum of the square of the roots and the sum of the reciprocals of the roots of
2x
3
− x + 2 = 0 .
Solution: Let a , b, c be the roots of 2x
3
− x + 2 = 0. From the Viète Formulæ the sum of the roots is a + b + c = −
2 = 0
and the sum of the roots taken two at a time is ab + ac + bc =
−1 2
. To find a
2
+ b
2
+ c
2
we observe that a
2
+ b
2
+ c
2
= a + b + c
2
− 2ab + ac + bc .
Hence a
2
+ b
2
+ c
2
= 0
2
− 2−1 2 = 1.
Finally, as abc = −2 2 = −1, we gather that
1 a
+ 1
b +
1 c
= ab + ac + bc
abc =
−1 2
−1 = 1
2.
714 Example Let
α
,
β
,
γ
be the roots of x
3
− x
2
+ 1 = 0. Find 1
α
2
+ 1
β
2
+ 1
γ
2
.
Viète’s Formulae 109
Solution: From x
3
− x
2
+ 1 = 0 we deduce that 1 x
2
= 1 − x. Hence 1
α
2
+ 1
β
2
+ 1
γ
2
= 1 −
α
+ 1 −
β
+ 1 −
γ
= 3 −
α
+
β
+
γ
= 3 − 1 = 2 .
Together with the Viète Formulæ we also have the Newton-Girard Identities for the sum of the powers s
k
=
α
k 1
+
α
k 2
+ ··· +
α
k n
of the roots: a
s
1
+ a
1
= 0 ,
a s
2
+ a
1
s
1
+ 2a
2
= 0 ,
a s
3
+ a
1
s
2
+ a
2
s
1
+ 3a
3
= 0 ,
etc..
715 Example If a
, b, c are the roots of x
3
− x
2
+ 2 = 0, find a
2
+ b
2
+ c
2
a
3
+ b
3
+ c
3
and a
4
+ b
4
+ c
4
. Solution: First observe that
a
2
+ b
2
+ c
2
= a + b + c
2
− 2ab + ac + bc = 1
2
− 20 = 1 .
As x
3
= x
2
− 2 , we gather
a
3
+ b
3
+ c
3
= a
2
− 2 + b
2
− 2 + c
2
− 2 = a
2
+ b
2
+ c
2
− 6 = 1 − 6 = −5 .
Finally, from x
3
= x
2
− 2 we obtain x
4
= x
3
− 2x, whence a
4
+ b
4
+ c
4
= a
3
− 2a + b
3
− 2b + c
3
− 2c = a
3
+ b
3
+ c
3
− 2a + b + c = −5 − 21 = −7 .
716 Example USAMO 1973 Find all solutions real or complex of the system
x + y + z = 3 ,
x
2
+ y
2
+ z
2
= 3 ,
x
3
+ y
3
+ z
3
= 3 .
Solution: Let x , y, z be the roots of
pt = t − xt − yt − z = t
3
− x + y + zt
2
+ xy + yz + zxt − xyz .
Now xy + yz + zx = x + y + z
2
2 − x
2
+ y
2
+ z
2
2 = 92 − 32 = 3 and from x
3
+ y
3
+ z
3
− 3xyz = x + y + zx
2
+ y
2
+ z
2
− xy − yz − zx we gather that xyz = 1
. Hence pt = t
3
− 3t
2
+ 3t − 1 = t − 1
3
. Thus x = y = z = 1 is the only solution of the given system.
110 Chapter 6
Practice
717 Problem Suppose that
x
n
+ a
1
x
n−1
+ a
2
x
n−2
+ ··· + a
n
= x + r
1
x + r
2
··· x + r
n
where r
1
, r
2
, . . . ,r
n
are real numbers. Shew that n − 1a
2 1
≥ 2na
2
.
718 Problem USAMO 1984
The product of the roots of x
4
− 18x
3
+ kx
2
+ 200x − 1984 = 0 is −32. Determine k.
719 Problem The equation x
4
− 16x
3
+ 94x
2
+ px + q = 0 has two double roots. Find p + q
.
720 Problem If
α
1
,
α
2
, . . . ,
α
100
are the roots of x
100
− 10x + 10 = 0 ,
find the sum
α
100 1
+
α
100 2
+ ··· +
α
100 100
.
721 Problem
Let
α
,
β
,
γ
be the roots of x
3
− x − 1 = 0 . Find
1
α
3
+ 1
β
3
+ 1
γ
3
y
α
5
+
β
5
+
γ
5
.
722 Problem The real numbers
α
,
β
satisfy
α
3
− 3
α
2
+ 5
α
− 17 = 0 ,
β
3
− 3
β
2
+ 5
β
+ 11 = 0 .
Find
α
+
β
.
6.5 Lagrange’s Interpolation
