116
The middle equation gives
a
(
c
−
b
)
=
0, so that either
a
=
0 or
b
=
c
. In the
fi
rst case, we obtain
c
+
b
=
10
cb
=
1
.
Substituting
c
=
b
−
1
, the
fi
rst equation gives
b
2
−
10
b
+
1
=
0.
But the quadratic formula gives
b
=
5
±
2
√
6, which is irrational.
On the other hand,if
b
=
c
, then
bc
=
1 implies
b
= ±
1
=
c
. The
fi
rst equation gives
a
2
= −
10
±
2
<
0, and this is also impossible.
We conclude that there is no factorization of
f
(
x
)
in
Q
[
x
]
.
(ix)
f
(
x
)
=
x
6
−
210
x
−
616.
Solution.
Eisenstein
’
s criterion applies, for 7

210 and 7

616,
but 7
2
616.
(x)
f
(
x
)
=
350
x
3
+
x
2
+
4
x
+
1.
Solution.
Reducing mod 3 to gives an irreducible cubic in
F
3
[
x
]
.
3.88
If
p
is a prime, prove that there are exactly
1
3
(
p
3
−
p
)
monic irreducible
cubic polynomials in
F
p
[
x
]
.
Solution.
Absent.
3.89
Prove that there are exactly 6 irreducible quintics in
F
2
[
x
]
.
Solution.
There are 32 quintics in
F
2
[
x
]
, 16 of which have constant term 0;
that is, have 0 as a root. Of the 16 remaining polynomials, we may discard
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 Fall '11
 KeithCornell
 Quadratic equation, irreducible polynomials

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