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Standard VI
Maths
Question
The output of
z
3
+
2
z
2
+
5
z
+
1
, where
z
=
−
1
−
1
−
2
−
3
None of the above
A
−
2
B
−
1
C
−
3
D
None of the above
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Solution
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z
3
+
2
z
2
+
5
z
+
1
=
(
−
1
)
3
+
2
×
(
−
1
)
2
+
5
×
(
−
1
)
+
1
=
−
1
+
2
−
5
+
1
=
−
3
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