144
(ii)
Show that one root of
f
(
X
)
=
X
3
+
X
2
−
36 is an integer and
fi
nd the other two roots. Compare your method with Cardano
’
s
formula and with Vi`ete
’
s trigonometric solution.
Solution.
Corollary 3.91 says that any integer root is a divisor of
36, and one checks that 3 is a root; therefore,
X
3
+
X
2
−
36
=
(
X
−
3
)(
X
2
+
4
X
+
12
).
The quadratic formula gives the other two roots:
−
4
±
√
8, both
of which are negative.
The other two methods are longer. Both require the substitution
X
=
x
−
1
3
, yielding the reduced cubic
x
3
−
1
3
x
−
970
27
. We do not
give the calculations.
5.6
Show that if
u
is a root of a polynomial
f
(
x
)
∈
R
[
x
]
, then the complex
conjugate
u
is also a root of
f
(
x
)
.
Solution.
Let
f
(
x
)
=
∑
r
i
x
i
. If 0
=
∑
r
i
u
i
, then since complex conju
gation is an isomorphism
fi
xing each real number,
0
=
0
=
r
i
u
i
=
r
i
u
i
=
f
(
u
).
5.7
Assume that 0
≤
3
α <
360
◦
.
(i)
If cos 3
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 Fall '11
 KeithCornell
 Cos, Quadratic equation, Ri

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