Math, asked by harshita9752, 2 days ago

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

Answered by jayp32068
0

Step-by-step explanation:

Let p(x)=x³+ax²+bx+6

i ) it is given that , (x-2) is a factor of p(x) ,then

p(x)=0

=> 2³+a(2)²+b×2+6=0

=> 8+4a+2b+6=0

=> 4a+2b+14=0

Divide each term by 2 , we get

=> 2a+b+7=0 ---(1)

ii)It is given that, if p(x) divided by (x-3) leaves a remainder 3

=> p(3)=3

=> 3³+a(3)²+b×3+6=3

=> 27+9a+3b+6-3=0

=> 9a+3b+30=0

Divide each term by 3 , we get

=> 3a+b+10=0 ---(2)

Subtract equation (1) from equation (2) , we get

a + 3 = 0

=> a = -3

Now,

Substitute a=-3 in equation (1) , we get

2(-3)+b+7=0

=> -6+b+7=0

=> b+1=0

=> b = -1

Therefore,

a=-3 and b=-1

Answered by devanshu1234321
26

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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