Question

one factor of x^4+x^2+1

Answer 1

Consider x4 + x2 + 1 = (x4 + 2x2 + 1) – x2 = [(x2)2 + 2x2 + 1] – x2 = [x2 + 1]2 – x2 It is in the form of (a2 – b2) = (a + b)(a – b) Hence [x2 + 1]2 – x2 = [x2 + 1 + x] [x2 + 1 – x]

Answer 2

Answer:
x
4
+
x
2
+
1
=
(
x
2
−
x
+
1
)
(
x
2
+
x
+
1
)
Explanation:
This really makes a bit more sense in the complex numbers...
First note that:
(
x
2
−
1
)
(
x
4
+
x
2
+
1
)
=
x
6
−
1

So zeros of
x
4
+
x
2
+
1
are also zeros of
x
6
−
1
.
What are the zeros of
x
6
−
1
?
The real zeros are
1
and
−
1
, which are zeros of
x
2
−
1
, the factor we introduced. So the four zeros of
x
4
+
x
2
+
1
are the four complex zeros of
x
6
−
1
apart from
±
1
.
Here are all
6
in the complex plane:
graph{((x-1)^2+(y-0)^2-0.002)((x-1/2)^2+(y-sqrt(3)/2)^2-0.002)((x+1/2)^2+(y-sqrt(3)/2)^2-0.002)((x+1)^2+(y-0)^2-0.002)((x+1/2)^2+(y+sqrt(3)/2)^2-0.002)((x-1/2)^2+(y+sqrt(3)/2)^2-0.002) = 0 [-2.5, 2.5, -1.25, 1.25]}
They form the vertices of a regular hexagon.
de Moivre's formula tells us that:
(
cos
θ
+
i
sin
θ
)
n
=
cos
n
θ
+
i
sin
θ

where
i
is the imaginary unit, satisfying
i
2
=
−
1

For instance we find:
(
cos
(
π
3
)
+
i
sin
(
π
3
)
)
6
=
cos
2
π
+
i
sin
2
π
=
1
+
0
=
1
That's the zero
1
2
+
√
3
2
i
that we see in Q1.
If
a
is a zero of a polynomial then
(
x
−
a
)
is a factor.
Hence:
x
4
+
x
2
+
1
=
(
x
−
1
2
−
√
3
2
i
)
(
x
−
1
2
+
√
3
2
i
)
(
x
+
1
2
−
√
3
2
i
)
(
x
+
1
2
+
√
3
2
i
)
x
4
+
x
2
+
1
=
(
x
2
−
x
+
1
)
(
x
2
+
x
+
1
)