Alg5.1.10

Write $\frac{x^3}{(2x-1)(x+2)(x-3)}\,$ as the sum of partial fractions.

Since the given fraction is an improper fraction,first do the division to change it into a proper fraction.

Note that the quotient is $\frac{1}{2}\,$ (See the greatest degree terms in the numerator and denominator).

Therefore the given fraction can be written as $\frac{1}{2}+\frac{A}{2x-1}+\frac{B}{x+2}+\frac{C}{x-3}\,$

Simplifying the equations we get

$x^3=(2x-1)(x+2)(x-3)+2A(x+2)(x-3)+2B(2x-1)(x-3)+2C(2x-1)(x+2)\,$

To find A,B,C substitute $x=\frac{1}{2}\,$

We get $\frac{1}{4}=2A(\frac{1}{2}+2)(\frac{1}{2}-3)\,$

$\frac{1}{4}=2A(-\frac{25}{4})=A(\frac{-25}{2}),A=-\frac{1}{50}\,$

Substituting $x=-2\,$ we get

$-16=2B(2(-2)-1)(-2-3)=50B,B=-\frac{8}{25}\,$

Substituting $x=3\,$ we get

$54=2C(2(3)-1)(3+2)=50C,C=\frac{27}{25}\,$

Therefore $\frac{x^3}{(2x-1)(x+2)(x-3)}=\frac{1}{2}-\frac{1}{50(2x-1)}-\frac{8}{25(x+2)}+\frac{27}{25(x-3)}\,$