X³ + X² + x + 1 is*
Answers
Answer:
You can factor by grouping to find:
x
3
+
x
2
−
x
−
1
=
(
x
2
−
1
)
(
x
+
1
)
Then use the difference of squares identity to find:
(
x
2
−
1
)
(
x
+
1
)
=
(
x
−
1
)
(
x
+
1
)
(
x
+
1
)
=
(
x
−
1
)
(
x
+
1
)
2
Explanation:
First factor by grouping:
x
3
+
x
2
−
x
−
1
=
(
x
3
+
x
2
)
−
(
x
+
1
)
=
x
2
(
x
+
1
)
−
1
(
x
+
1
)
=
(
x
2
−
1
)
(
x
+
1
)
Then notice that
x
2
−
1
=
x
2
−
1
2
is a difference of squares, so we can use the difference of squares identity [
a
2
−
b
2
=
(
a
−
b
)
(
a
+
b
)
] to find:
(
x
2
−
1
)
(
x
+
1
)
=
(
x
−
1
)
(
x
+
1
)
(
x
+
1
)
=
(
x
−
1
)
(
x
+
1
)
2
Alternatively, notice that the sum of the coefficients (
1
+
1
−
1
−
1
) is
0
, so
x
=
1
is a zero of this cubic polynomial and
(
x
−
1
)
is a factor.
Divide
x
3
+
x
2
−
x
−
1
by
(
x
−
1
)
to get
x
2
+
2
x
+
1
:
enter image source here
Then recognise that
x
2
+
2
x
+
1
=
(
x
+
1
)
2
is a perfect square trinomial. One little trick to spot this one is that
11
2
=
121
, the
1
,
2
,
1
matching the coefficients of the quadratic and
1
,
1
matching the coefficients of the linear factor
Answer:
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