If x 4 ( x - 1 ) ( x - 2 ) = f ( x ) + A x - 1 + B x - 2 , then

Given $\frac{x^4}{(x-1)(x-2)}=f\left(x\right)+\frac{A}{x-1}+\frac{B}{x-2}$

$\Rightarrow \frac{x^4}{(x-1)(x-2)}=\frac{f\left(x\right)\left(x-1\right)\left(x-2\right)+A\left(x-2\right)+B\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}$

$\Rightarrow x^4=f\left(x\right)\left(x-1\right)\left(x-2\right)+A\left(x-2\right)+B\left(x-1\right)$

On putting $x=1$, we get $A=-1$

On putting $x=2$, we get $B=16$

Then, $x^4=f\left(x\right)\left(x-1\right)\left(x-2\right)-\left(x-2\right)+16\left(x-1\right)$

$\Rightarrow x^4-16x+16+x-2=f\left(x\right)\left(x-1\right)\left(x-2\right)$

$\Rightarrow f\left(x\right)=\frac{x^4-15x+14}{\left(x-1\right)\left(x-2\right)}$

On adding and Subtracting $x^2$, we get

$f\left(x\right)=\frac{x^4-15x+14+x^2-x^2}{\left(x-1\right)\left(x-2\right)}=\frac{\left(x^4-x^2-12x+12\right)+\left(x^2-3x+2\right)}{\left(x^2-3x+2\right)}$

$\Rightarrow f\left(x\right)=\frac{\left(x^4-x^2-12x+12\right)}{\left(x^2-3x+2\right)}+1$

$\Rightarrow f\left(x\right)=\frac{\left(x-1\right)\left(x^3+x^2-12\right)}{\left(x-1\right)\left(x-2\right)}+1$

$\Rightarrow f\left(x\right)=\frac{x^3+x^2+x-14}{x-2}$

$\Rightarrow f\left(x\right)=x^2+3x+7$