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2018 CHMMC (Fall)
5
2018 CHMMC Tiebreaker 5 - roots of p(x) = 2x^5 - 3x^3 + 2x -7
2018 CHMMC Tiebreaker 5 - roots of p(x) = 2x^5 - 3x^3 + 2x -7
Source:
March 2, 2024
algebra
polynomial
CHMMC
Problem Statement
Let
a
,
b
,
c
,
d
,
e
a,b, c, d,e
a
,
b
,
c
,
d
,
e
be the roots of
p
(
x
)
=
2
x
5
−
3
x
3
+
2
x
−
7
p(x) = 2x^5 - 3x^3 + 2x -7
p
(
x
)
=
2
x
5
−
3
x
3
+
2
x
−
7
. Find the value of
(
a
3
−
1
)
(
b
3
−
1
)
(
c
3
−
1
)
(
d
3
−
1
)
(
e
3
−
1
)
.
(a^3 - 1)(b^3 - 1)(c^3 - 1)(d^3 - 1)(e^3 - 1).
(
a
3
−
1
)
(
b
3
−
1
)
(
c
3
−
1
)
(
d
3
−
1
)
(
e
3
−
1
)
.
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