x^9-1 factorise this question plzz
Answers
Step-by-step explanation:
The easiest way to do this is probably using Complex arithmetic and de Moivre's formula:
(
cos
θ
+
i
sin
θ
)
n
=
cos
n
θ
+
i
sin
n
θ
We find that the nine
9
th roots of
−
1
are:
cos
(
k
π
9
)
+
i
sin
(
k
π
9
)
for
k
=
±
1
,
±
3
,
±
5
,
±
7
,
9
In order to find factors with Real coefficients, we can pair up the Complex conjugate pairs like this:
(
x
−
cos
(
k
π
9
)
−
i
sin
(
k
π
9
)
)
(
x
−
cos
(
−
k
π
9
)
−
i
sin
(
−
k
π
9
)
)
=
(
x
−
cos
(
k
π
9
)
−
i
sin
(
k
π
9
)
)
(
x
−
cos
(
k
π
9
)
+
i
sin
(
k
π
9
)
)
=
(
x
−
cos
(
k
π
9
)
)
2
−
(
i
sin
(
k
π
9
)
)
2
=
x
2
−
2
cos
(
k
π
9
)
x
+
1
for
k
=
1
,
3
,
5
,
7
Hence:
x
9
+
1
=
(
x
+
1
)
(
x
2
−
2
cos
(
π
9
)
x
+
1
)
(
x
2
−
2
cos
(
3
π
9
)
x
+
1
)
(
x
2
−
2
cos
(
5
π
9
)
x
+
1
)
(
x
2
−
2
cos
(
7
π
9
)
x
+
1
)