146
5.13
Find the roots of
x
3
−
6
x
+
4.
Solution.
We have
q
=−
6,
r
=
4,
D
=−
16, and
α
3
=−
2
+
2
i
.UseDe
Moivre
’
s theorem to
f
nd cube roots of
−
2
+
2
i
:
(
1
+
i
)
3
=−
2
+
2
i
.
Hence,
α
=
1
+
i
,
β
=
1
−
i
, and a root of the polynomial is
α
+
β
=
2.
The other two roots can be found by two methods. We can use the cubic
formula to obtain:
ω(
1
+
i
)
+
ω
2
(
1
−
i
)
and
ω
2
(
1
+
i
)
+
ω(
1
−
i
),
or one can use the division algorithm and the quadratic formula:
x
3
−
6
x
+
4
=
(
x
−
2
)(
x
2
+
2
x
−
2
),
so that the other roots are
−
1
±
√
3. It is not clear from the cubic for
mula that the roots are all real; however, 27
r
2
+
4
q
3
=−
432
<
0, and
Proposition 5.9 applies.)
5.14
Find the roots of
x
4
−
15
x
2
−
20
x
−
6.
Solution.
This is a realistic problem; the implementation of the formula is
rather long (but see the remark at the end of this solution).
Set
Q
=−
15,
R
=−
20, and
S
=−
6. Then
j
6
+
2
Qj
4
+
(
Q
2
−
4
S
)
j
2
−
R
2
=
j
6
−
30
j
4
+
249
j
2
−
400
.
We now make this cubic (in
j
2
) reduced with the substitution
j
2
=
y
+
10,
obtaining
F
(
y
)
=
y
3
−
51
y
+
90
.
Apply the cubic formula:
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 Fall '11
 KeithCornell
 Quadratic equation, Complex number, cubic formula

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