If f (x) = x^4 - 2 x^3 - 3 x^2 - ax + b is divided by x-1 and x--1 the remainders are 5 and 19 respectively find the value of a and b

On dividing, f (x) by (x-1) we get remainder = b-a-4

And, we are given the remainder is 5

So, b-a-4=5

And, On dividing f (x) by (x+1) we get remainder=b+a

So, b+a=19

As, b-a-4=5 And b+a=19

To solve for a and b let us add the two equations

So, 2b+0a-4=5+19

Or. 2b-4=24

Adding 4 on both sides we get

2b-4+4=24+4

2b=28

Divide by 2 on both sides

2b/2=28/2

b=14

As, a+b=19 and b=14

So, a+14=19

Subtracting 14 from both sides, we get

a+14-14=19-14

a+0=5

a=5

So, a=5 and b=14