plzz solve this question fast I will mark you as brainlist
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Answer:
no it is not
Step-by-step explanation:
samson99:
plzz give the solution step by step
(
x
−
a
)
is a factor of
f
(
x
)
⇔
f
(
a
)
=
0
f
(
x
)
=
4
x
3
−
3
x
2
−
8
x
+
4
we have
(
x
−
2
)
⇒
a
=
2
f
(
2
)
=
4
×
2
3
−
3
×
2
2
−
8
×
2
+
4
f
(
2
)
=
4
×
8
−
3
×
4
−
16
+
4
f
(
2
)
=
32
−
12
−
16
+
4
f
(
2
)
=
8
≠
0
∴
x
−
2
is not a factor of
4
x
3
−
3
x
2
−
8
x
+
4
however by the remainder theorem when
f
(
x
)
=
4
x
3
−
3
x
2
−
8
x
+
4
is divided by
(
x
−
2
)
the remainder is
8
the factor theorem being a special case of the remainder theorem
x
−
a
)
is a factor of
f
(
x
)
⇔
f
(
a
)
=
0
f
(
x
)
=
4
x
3
−
3
x
2
−
8
x
+
4
we have
(
x
−
2
)
⇒
a
=
2
f
(
2
)
=
4
×
2
3
−
3
×
2
2
−
8
×
2
+
4
f
(
2
)
=
4
×
8
−
3
×
4
−
16
+
4
f
(
2
)
=
32
−
12
−
16
+
4
f
(
2
)
=
8
≠
0
∴
x
−
2
is not a factor of
4
x
3
−
3
x
2
−
8
x
+
4
however by the remainder theorem when
f
(
x
)
=
4
x
3
−
3
x
2
−
8
x
+
4
is divided by
(
x
−
2
)
the remainder is
8
the factor theorem being a special case of the remainder theorem
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