Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
The factorization of 128-32 x+2 x2 is
Q. The factorization of
128
−
32
x
+
2
x
2
is
49
150
Polynomials, LCM and HCF of Polynomials
Report Error
A
2
(
x
−
8
)
B
2
(
x
−
8
)
2
C
(
x
−
8
)
(
x
+
8
)
D
2
(
x
−
8
)
(
x
+
8
)
Solution:
128
−
32
x
+
2
x
2
=
2
x
2
−
32
x
+
128
=
0
=
2
x
2
−
16
x
−
16
x
+
128
=
0
=
2
x
(
x
−
8
)
−
16
(
x
−
8
)
=
0
\
=
(
2
x
−
16
)
(
x
−
8
)
=
2
(
x
−
8
)
2