If `f(x)=x^4-2x^3+3x^2-a x+b` is a polynomial such that when1 it is divided by `x-1` and `x+1` , remainders are `5` and `19` respectively.

Explanation

`x^2 2x - 3`
`= x^2+ 3x - x - 3`
`= x(x + 3) - (x + 3)`
`= (x + 3)(x - 1)`
Now, by remainder theorem, remainder `= 0`
Taking, `x + 3` as the factor
So, `x = -3`
`x^4 + 2x^3 - 2x^2+ 2x - 3 = 0`
`=>(-3)^4 + 2(3)^3 - 2(-3)^2 + 2(-3) - 3 = 0`
`=>81 - 54 - 18 - 6 - 3 = 0`
`=>81 - 54 - 27 = 0`
`=>81 - 81 = 0`
`=>0 = 0`
Hence, `(x + 3)` is the factor of given equation,
Now, checking for `(x - 1)` as a factor,
So, `x = 1`
`=>x^4 + 2x^3 - 2x^2 + 2x - 3 = 0`
`=>(1)⁴^4+ 2(1)^3 - 2(1)^2 + 2(1) - 3 = 0`
`=>1 + 2 - 2 + 2 - 3 = 0`
`=>3 - 3 = 0`
`=>0 = 0`
Then,` x - 1` is Also a factor,
As both `(x - 1)` and `(x + 3)` are factors, we can say that the given equation is divisible by `x^2 +2x - 3`.