if x³+ax²+bx+4 is divided by x-2, the remainder is 6. if it is divided by x+1, the remainder is -3. find a and b.
Answers
Answer:
Answer
f(x)=x
3
+ax
2
+bx+6
(x−2) in a factor ⇒f(2)=0
f(2)⇒2
3
+a(2)
2
+2b+6=0
⇒8+4a+2b+b=0
⇒4a+2b+14=0
⇒2a+b+7=0 -(1)
f(x−3)=3 (Remainder)
⇒f(3)⇒3
3
+a(3)
2
+b×3+6=3
⇒27+9a+3b+b=3
⇒9a+3b+30=0
⇒3a+b+10=0 _____ (2)
From (1) & (2)
b=−2a−7 & b=−10−3a
−2a−7=−10−3a
3a−2a=−10+7
a=−3
from (3)
b=−2(a)−7=2(−3)−7
=6−7
b=−1
QUESTION:-
if x³+ax²+bx+4 is divided by x-2, the remainder is 6. if it is divided by x+1, the remainder is -3. find a and b.
EXPLANATION:-
Let,
p(x)=x³+ax²+bx+4
Let's use remainder theorem here,
x-2=0
x=2
Now acc. to the remainder theorem ,
p(2)=6 (It is given that when x-2 is divided by the polynomial then the remainder is 0)
p(2)= (2)³+a(2)²+b(2)+4
p(2)=8+4a+2b+4
p(2)=12+4a+2b
p(2)=2(6+2a+b)
Since p(2)=6,
6=2(6+2a+b)
6/2=6+2a+b
3=6+2a+b
-3=2a+b ------[EQ-1]
Now similarily we can write:-
x+1=0
x=-1
So,
p(-1)=-3 (SAME REASON)
p(-1)=(-1)³+a(-1)²+b(-1)+4
p(-1)=-1+a-b+4
p(-1)=3+a-b
p(-1)=-3
-3=3+a-b
-6=a-b --------[EQ-2]
From eq-1 and 2 we have,
-3=2a+b
-3-2a=b
Put -3-2a=b in the 2 eq
6=a-b
6=a-(-3-2a)
6=a+3+2a
6=3a+3
3=3a
a=1
So the value of a is 1 ,now put a=1 in eq-2
-3=2a+b
-3=2+b
-3-2=b
b=-5
Thus,
a=1 and b=-5
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