if z=-10, find z^3+2z^2-7z-10
Answers
Answered by
0
Step-by-step explanation:
Given:
z
4
−
2
z
3
+
7
z
2
−
4
z
+
10
=
0
If
z
=
a
i
for some real number
a
then, the terms of even degree are real and those of odd degree imaginary...
0
=
(
a
i
)
4
−
2
(
a
i
)
3
+
7
(
a
i
)
2
−
4
(
a
i
)
+
10
0
=
(
a
4
−
7
a
2
+
10
)
+
2
a
(
a
2
−
2
)
i
0
=
(
a
2
−
5
)
(
a
2
−
2
)
+
2
a
(
a
2
−
2
)
i
0
=
(
(
a
2
−
5
)
+
2
a
i
)
(
a
2
−
2
)
Hence
a
=
±
√
2
So two of the roots of the original quartic are
±
√
2
i
, with associated factors:
(
z
−
√
2
i
)
(
z
+
√
2
i
)
=
z
2
+
2
We find:
z
4
−
2
x
3
+
7
z
2
−
4
z
+
10
=
(
z
2
+
2
)
(
z
2
−
2
z
+
5
)
z
4
−
2
x
3
+
7
z
2
−
4
z
+
10
=
(
z
2
+
2
)
(
z
2
−
2
z
+
1
+
4
)
z
4
−
2
x
3
+
7
z
2
−
4
z
+
10
=
(
z
2
+
2
)
(
(
z
−
1
)
2
−
(
2
i
)
2
)
z
4
−
2
x
3
+
7
z
2
−
4
z
+
10
=
(
z
2
+
2
)
(
z
−
1
−
2
i
)
(
z
−
1
+
2
i
)
So the other two roots are:
z
=
1
±
2
i
Answer
Answered by
2
Answer:
- 740
Step-by-step explanation:
z³ + 2z² - 7z - 10
= ( - 10 )³ + 2 ( - 10 )² - 7 ( - 10 ) - 10
= - 1000 + 2 ( 100 ) + 70 - 10
= - 1000 + 200 + 60
= - 1000 + 260
= - 740
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