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Byju's Answer
Standard IX
Mathematics
Taking Out Common Factors
Factorize: ...
Question
Factorize:
a
2
+
1
a
2
−
2
−
3
a
+
3
a
A
(
a
−
6
a
)
(
a
−
1
a
−
1
)
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B
(
a
−
3
a
)
(
a
−
1
a
−
9
)
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C
(
a
−
2
a
)
(
a
−
1
a
−
7
)
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D
(
a
−
1
a
)
(
a
−
1
a
−
3
)
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Solution
The correct option is
D
(
a
−
1
a
)
(
a
−
1
a
−
3
)
Given,
a
2
+
1
a
2
−
2
−
3
a
+
3
a
=
a
2
+
1
a
2
−
2
(
a
)
(
1
a
)
−
3
(
a
−
1
a
)
=
(
a
−
1
a
)
2
−
3
(
a
−
1
a
)
=
(
a
−
1
a
)
(
a
−
1
a
−
3
)
Suggest Corrections
2
Similar questions
Q.
If
f
(
x
)
=
[
a
+
x
1
+
x
]
a
+
1
+
2
x
then
a
a
+
1
[
2
l
o
g
a
+
1
−
a
2
a
]
is
Q.
Given
a
1
=
1
2
a
0
+
A
a
0
,
a
2
=
1
2
a
1
+
A
a
1
and
a
n
+
1
=
1
2
a
n
+
A
a
n
for
n
≥
2
,
where
a
>
0
,
A
>
0
.
Prove
that
a
n
-
A
a
n
+
A
=
a
1
-
A
a
1
+
A
2
n
-
1
.
Q.
If
A
2
=
3
A
−
2
I
and
A
8
=
a
A
+
b
I
, then
a
+
b
is
Q.
Find the inverse of the matrix
A
=
a
b
c
1
+
b
c
a
and show that
a
A
-
1
=
a
2
+
b
c
+
1
I
-
a
A
.
Q.
Factorize:
24
a
3
+
37
a
2
−
5
a
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