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.​

Answer 1

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

Hope it's helps u❤✨

Answer 2

Answer:

If we put the value of a= -3 and b= 3 in both the conditions then we can get the answer.