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

When $f\left(x\right)$  is divided by $x-1andx+1$  the remainders are 5 and 19 respectively.

$\therefore$  $f\left(1\right)=5andf\left(-1\right)=19$

$\Rightarrow 1^4-2\times 1^3+3\times 1^2-a\times 1+b=5$

And, ${\left(-1\right)}^4-2\times {\left(-1\right)}^3+3\times {\left(-1\right)}^2-a\times -1+b=19$

$\Rightarrow 1-2+3-a+b=5and1+2+3+a+b=19$

$\Rightarrow 2-a+b=5and6+a+b=19\Rightarrow -a+b=3anda+b=13$

Adding these two equations, we get

$(-a+b)+(a+b)=3+13\Rightarrow 2b=16\Rightarrow b=8$  

Putting $b=8$  in$-a+b=3,$  we get

$-a+8=3\Rightarrow a=5$

Putting the values of $aandb$ in

$f\left(x\right)=x^4-2x^3+3x^2-5x+8$

The remainder when $f\left(x\right)$  is divided by $\left(x-2\right)$  is equal to $f\left(2\right).$

So, Remainder $=f\left(2\right)=2^4-2\times 2^3+3\times 2^2-5\times 2+8=16-16+12-10+8=10$