Answer:
There are four questions in your one question.
Let us take
(
B
)
(
x
3
−
4
x
2
−
2
x
+
5
)
÷
(
x
−
1
)
Here,
p
(
x
)
=
x
3
−
4
x
2
−
2
x
+
5
and
divisor
x
=
1
We take coefficients of
p
(
x
)
→
1
,
−
4
,
−
2
,
5
and set the problem as shown below.
Put zero {0} below the first number {1} and add :
1
+
0
=
1
.
1
∣
1
...
.
.
−
4
...
.
.
−
2
...
...
5
.2
−
∣
∣
∣
0
...
...
...
...
...
...
...
...
...
...
...
.
−−−−−−−−−−−−−−−−−−−−
...
.
1
Now multiply this {1}with divisor
(
1
)
→
1
×
(
1
)
=
1
and put below second number{-4} and add
→
−
4
+
1
=
−
3
.
1
∣
1
...
.
.
−
4
...
.
.
−
2
...
...
.
5
−
∣
∣
∣
0
...
...
...
.
1
...
...
...
...
...
...
...
.
−−−−−−−−−−−−−−−−−−−−−
...
.
1
...
...
.
−
3
...
.
.20
...
...
...
|
0
|
−−
Again repeat the process :
i
.
e
.
−
3
×
(
1
)
=
−
3
and
−
2
+
(
−
3
)
=
−
5
.
1
∣
1
...
...
...
−
4
...
...
.
−
2
...
...
.
5
−
∣
∣
∣
0
...
...
...
.
.
1
...
...
...
.
−
3
...
...
...
10
−−−−−−−−−−−−−−−−−−−−−−−−−
...
.
1
...
...
...
−
3
...
...
...
−
5
...
...
...
|
0
|
−−
Again ,
−
5
×
(
1
)
=
−
5
and
(
5
)
+
(
−
5
)
=
0
.
1
∣
1
...
...
.
−
4
...
...
.
−
2
...
...
.
.
5
−
∣
∣
∣
0
...
...
...
.
.
1
...
...
.
−
3
...
...
−
5
−−−−−−−−−−−−−−−−−−−−−−−
...
.
1
...
...
.
−
3
...
...
.
−
5
...
...
...
|
0
|
−−
We can see that , quotient polynomial :
q
(
x
)
=
x
2
−
3
x
−
5
and
the Remainder
=
0
Hence ,
(
x
3
−
4
x
2
−
2
x
+
5
)
=
(
x
−
1
)
(
x
2
−
3
x
−
5
)
+
0
Note: For
(
A
)
,
(
C
)
,
and
(
D
)
Step-by-step explanation:
welcome
