2 Consider a polynomial, f(x) = ax3 + bx² + x + 3. If x + 3 is a factor of f(x) and if f(x) is divided by x + 2, then we get remainder as 5. Then, find the values of a and b.
Answers
Answer:
a &b =0&2
Step-by-step explanation:
Let p(x)=ax
3
+bx
2
+x−6
Since (x+2) is a factor of p(x), then by Factor theorem p(−2)=0
⇒a(−2)
3
+b(−2)
2
+(−2)−6=0
⇒−8a+4b−8=0
⇒−2a+b=2 ...(i)
Also when p(x) is divided by (x-2) the remainder is 4, therefore by Remainder theorem p(2)=4
⇒a(2)
3
+b(2)
2
+2−6=4
⇒8a+4b+2−6=4
⇒8a+4b=8
⇒2a+b=2 ...(ii)
Adding equation (i) and (ii), we get
(−2a+b)+(2a+b)=2+2
⇒2b=4⇒b=2
Putting b=2 in (i), we get
−2a+2=2
⇒−2a=0⇒a=0
Hence, a=0 and b=2
Step-by-step explanation:
f
(
x
)
=
a
x
3
+
b
x
2
+
x
+
2
3
∴
(
x
+
3
)
is factor of
f
(
x
)
.
By factor theorem,
f
(
−
3
)
=
0
⇒
a
(
−
3
)
3
+
b
(
−
3
)
2
+
(
−
3
)
+
2
3
=
0
⇒
−
27
a
+
9
b
−
7
3
=
0
⇒
−
27
a
+
9
b
=
7
3
⇒
3
(
−
9
a
+
3
b
)
=
7
3
⇒
−
9
a
+
3
b
=
7
9
∴
3
b
=
7
9
+
9
a
.......(1)
When
f
(
x
)
is divided by
(
x
+
2
)
, remainder is 4.
By remainder theorem,
f
(
−
2
)
=
4
⇒
a
(
−
2
)
3
+
b
(
−
2
)
2
+
(
−
2
)
+
2
3
=
4
⇒
−
8
a
+
4
b
−
4
3
=
4
⇒
−
24
a
+
12
b
−
4
=
12
⇒
−
24
a
+
12
b
=
16
⇒
−
6
a
+
3
b
=
4
∴
3
b
=
4
+
6
a
...........(2)
From (1) and (2), we get
3
b
=
7
9
+
9
a
and
3
b
=
6
a
+
4
7
9
+
9
a
=
6
a
+
4
⇒
9
a
−
6
a
=
4
−
7
9
⇒
3
a
=
29
9
∴
a
=
29
27
from (1), we get
3
b
=
7
9
+
9
a
=
7
9
+
9
×
29
27
⇒
3
b
=
7
+
87
9
⇒
3
b
=
94
9
∴
b
=
94
27