Q.Resolve $\dfrac{x^3}{(x-1)(x-2)}$ into partial fractions.

Step 1. Since $\deg(x^3)=3\ge\deg\big((x-1)(x-2)\big)=2$, the fraction is improper. Expand the denominator: $(x-1)(x-2)=x^2-3x+2$.

Step 2. Divide $x^3$ by $x^2-3x+2$: the first quotient term is $x$, since $x\cdot(x^2-3x+2)=x^3-3x^2+2x$; subtracting from $x^3$ leaves $3x^2-2x$.

Step 3. The next quotient term is $3$, since $3\cdot(x^2-3x+2)=3x^2-9x+6$; subtracting from $3x^2-2x$ leaves $7x-6$.

Step 4. So $x^3=(x^2-3x+2)(x+3)+(7x-6)$, i.e.

$\frac{x^3}{(x-1)(x-2)}=x+3+\frac{7x-6}{(x-1)(x-2)}.$ …

Improper fraction: polynomial long division first to obtain the integral part $Q(x)$, then resolve the proper r …

• Trying to apply the partial-fraction rules directly to the improper fraction without dividing first. …