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

Explanation

When `f(x)` is divided by `x-1` and `x+1` the remainder are `5` and `19` respectively.
`∴f(1)=5` and `f(−1)=19`
`⇒(1)^4−2×(1)^3+3×(1)^2−a×1+b=5` and `(−1)^4−2×(−1)^3+3×(−1)^2−a×(−1)+b=19`
`⇒1−2+3−a+b=5` and `1+2+3+a+b=19`
`⇒2−a+b=5` and `6+a+b=19`
`⇒−a+b=3` and `a+b=13`
Adding these two equations, we get:
`(−a+b)+(a+b)=3+13`
`⇒2b=16`
`⇒b=8`
Putting `b=8` and `−a+b=3`, we get:
`−a+8=3`
`⇒-a=−5`
`⇒a=5`
Putting the values of `a` and `b` in `f(x)=x^4−2x^3+3x^2−5x+8`
The remainder when `f(x)` is divided by `(x-2)` is equal to `f(2)`
So, Remainder `=f(2)=(2)^4−2×(2)^3+3×(2)^2−5×2+8=16−16+12−10+8=10`
`therefore` The remainder when `f(x)` is divided by `(x-2)=10`