Resolve the given fraction into partial fractions

$\frac{x^4}{(x-1)(x-2)}$

 $\left.x^2-3x+2\right)x^4$      $\left(x^2+3x+7\right.$

$\frac{x^4-3x^3+2x^2}{3x^3-2x^2}$

$\frac{3x^3-9x^2+6x}{7x^2-6x}$

$\frac{7x^2-21x+14}{15x-14}$

$\frac{x^4}{(x-1)(x-2)}=\frac{x^4}{x^2-3x+2}$

$x^2+3x+7+\frac{15x-14}{x^2-3x+2}$

$x^2+3x+7+\frac{15x-14}{x^2-3x+2}$

Let  $\frac{15x-14}{x^2-3x+2}=\frac{15x-14}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}$

$15x-14=A(x-2)+B(x-1)$.....(1)

Put$x=1$in (1) we get

$15\left(1\right)-14=A(1-2)+B\left(0\right)1=-A\Rightarrow \boxed{A=-1}$

Put $x=2$ in (1) we get

$15\left(2\right)-14=A\left(0\right)+B(2-1)$

$30-14=0+B$     $\Rightarrow \boxed{16=B}$

$\frac{x^4}{(x-1)(x-2)}=x^2+3x+7+\frac{(-1)}{x-1}+\frac{16}{x-2}$

$=x^2+3x+7-\frac{1}{x-1}+\frac{16}{x-2}$