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National and Regional Contests
Moldova Contests
Moldova Team Selection Test
2019 Moldova Team Selection Test
7
7
Part of
2019 Moldova Team Selection Test
Problems
(1)
another incredible polynomial
Source: Moldova TST 2019
3/10/2019
Let
P
(
X
)
=
a
2
n
+
1
X
2
n
+
1
+
a
2
n
X
2
n
+
.
.
.
+
a
1
X
+
a
0
P(X)=a_{2n+1}X^{2n+1}+a_{2n}X^{2n}+...+a_1X+a_0
P
(
X
)
=
a
2
n
+
1
X
2
n
+
1
+
a
2
n
X
2
n
+
...
+
a
1
X
+
a
0
be a polynomial with all positive coefficients. Prove that there exists a permutation
(
b
2
n
+
1
,
b
2
n
,
.
.
.
,
b
1
,
b
0
)
(b_{2n+1},b_{2n},...,b_1,b_0)
(
b
2
n
+
1
,
b
2
n
,
...
,
b
1
,
b
0
)
of numbers
(
a
2
n
+
1
,
a
2
n
,
.
.
.
,
a
1
,
a
0
)
(a_{2n+1},a_{2n},...,a_1,a_0)
(
a
2
n
+
1
,
a
2
n
,
...
,
a
1
,
a
0
)
such that the polynomial
Q
(
X
)
=
b
2
n
+
1
X
2
n
+
1
+
b
2
n
X
2
n
+
.
.
.
+
b
1
X
+
b
0
Q(X)=b_{2n+1}X^{2n+1}+b_{2n}X^{2n}+...+b_1X+b_0
Q
(
X
)
=
b
2
n
+
1
X
2
n
+
1
+
b
2
n
X
2
n
+
...
+
b
1
X
+
b
0
has exactly one real root.
algebra
polynomial