check if x+1 is a factored x^5-x^4+x^3-x^2-x-1
Answers
Answer:
Answer:
x
4
+
x
3
+
x
2
+
x
+
1
=
(
x
2
+
(
1
2
+
√
5
2
)
x
+
1
)
(
x
2
+
(
1
2
−
√
5
2
)
x
+
1
)
Explanation:
This quartic has four zeros, which are the non-Real Complex
5
th roots of
1
, as we can see from:
(
x
−
1
)
(
x
4
+
x
3
+
x
2
+
x
+
1
)
=
x
5
−
1
So if we wanted to factor this polynomial as a product of linear factors with Complex coefficients then we could write:
x
4
+
x
3
+
x
2
+
x
+
1
=
(
x
−
(
cos
(
2
π
5
)
+
i
sin
(
2
π
5
)
)
)
⋅
(
x
−
(
cos
(
4
π
5
)
+
i
sin
(
4
π
5
)
)
)
⋅
(
x
−
(
cos
(
6
π
5
)
+
i
sin
(
6
π
5
)
)
)
⋅
(
x
−
(
cos
(
8
π
5
)
+
i
sin
(
8
π
5
)
)
)
A cleaner algebraic approach is to notice that due to the symmetry of the coefficients, if
x
=
r
is a zero of
x
4
+
x
3
+
x
2
+
x
+
1
, then
x
=
1
r
is also a zero.
Hence there is a factorisation in the form:
x
4
+
x
3
+
x
2
+
x
+
1
=
(
x
−
r
1
)
(
x
−
1
r
1
)
(
x
−
r
2
)
(
x
−
1
r
2
)
=
(
x
2
−
(
r
1
+
1
r
1
)
x
+
1
)
(
x
2
−
(
r
2
+
1
r
2
)
x
+
1
)
So let's look for a factorisation:
x
4
+
x
3
+
x
2
+
x
+
1
=
(
x
2
+
a
x
+
1
)
(
x
2
+
b
x
+
1
)
=
x
4
+
(
a
+
b
)
x
3
+
(
2
+
a
b
)
x
2
+
(
a
+
b
)
x
+
1
Equating coefficients we find:
a
+
b
=
1
2
+
a
b
=
1
, so
a
b
=
−
1
and
b
=
−
1
a
Substituting
b
=
−
1
a
in
a
+
b
=
1
we get:
a
−
1
a
=
1
Hence:
a
2
−
a
−
1
=
0
Using the quadratic formula, we can deduce:
a
=
1
2
±
√
5
2
Since our derivation was symmetric in
a
and
b
, one of these roots can be used for
a
and the other for
b
, to find:
x
4
+
x
3
+
x
2
+
x
+
1
=
(
x
2
+
(
1
2
+
√
5
2
)
x
+
1
)
(
x
2
+
(
1
2
−
√
5
2
)
x
+
1
)
Answer is -4..............
