The problem asks to express the given rational expression:

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

in the partial fraction form:

$\frac{A}{3 - 2x} + \frac{B}{1 - x} + \frac{C}{(1 - x)^2}.$

Steps to Solve:

1. Set up the equation: $\frac{2x^2 - 3}{(3 - 2x)(1 - x)^2} = \frac{A}{3 - 2x} + \frac{B}{1 - x} + \frac{C}{(1 - x)^2}.$

2. Eliminate the denominators by multiplying through by $(3 - 2x)(1 - x)^2$: $2x^2 - 3 = A(1 - x)^2 + B(3 - 2x)(1 - x) + C(3 - 2x).$

3. Expand each term:

   • Expand $A(1 - x)^2$: $A(1 - x)^2 = A(1 - 2x + x^2) = A(x^2 - 2x + 1).$
   • Expand $B(3 - 2x)(1 - x)$: $B(3 - 2x)(1 - x) = B\big(3 - 3x - 2x + 2x^2\big) = B(2x^2 - 5x + 3).$
   • Expand $C(3 - 2x)$: $C(3 - 2x) = 3C - 2Cx.$

   So, the equation becomes: $2x^2 - 3 = A(x^2 - 2x + 1) + B(2x^2 - 5x + 3) + C(3 - 2x).$

4. Combine like terms: Collect terms involving $x^2$, $x$, and the constant terms: $2x^2 - 3 = (A + 2B)x^2 + (-2A - 5B - 2C)x + (A + 3B + 3C).$

5. Compare coefficients of $x^2$, $x$, and the constant term**: By comparing the left-hand side $2x^2 - 3$ with the right-hand side, we get:

   • Coefficient of $x^2$: $A + 2B = 2$,
   • Coefficient of $x$: $-2A - 5B - 2C = 0$,
   • Constant term: $A + 3B + 3C = -3$.

6. Solve the system of equations:

   • From $A + 2B = 2$, we can express $A = 2 - 2B$.

   • Substitute $A = 2 - 2B$ into the other two equations:

     • $-2(2 - 2B) - 5B - 2C = 0$: $-4 + 4B - 5B - 2C = 0 \implies -4 - B - 2C = 0 \implies B + 2C = -4.$
     • $(2 - 2B) + 3B + 3C = -3$: $2 - 2B + 3B + 3C = -3 \implies 2 + B + 3C = -3 \implies B + 3C = -5.$

     So, we now have a simpler system: $B + 2C = -4, \quad B + 3C = -5.$

   • Subtract the first equation from the second: $(B + 3C) - (B + 2C) = -5 - (-4) \implies C = -1.$

   • Substitute $C = -1$ into $B + 2C = -4$: $B + 2(-1) = -4 \implies B - 2 = -4 \implies B = -2.$

   • Substitute $B = -2$ into $A = 2 - 2B$: $A = 2 - 2(-2) = 2 + 4 = 6.$

7. Final Answer: The partial fraction decomposition is: $\frac{2x^2 - 3}{(3 - 2x)(1 - x)^2} = \frac{6}{3 - 2x} - \frac{2}{1 - x} - \frac{1}{(1 - x)^2}.$

Let me know if you'd like a detailed explanation of any specific step!

Related Questions:

1. How do partial fractions simplify solving integrals involving rational expressions?
2. Could we verify the result by recombining the fractions into a single term?
3. What happens if the denominator has repeated irreducible quadratic factors instead of linear ones?
4. How can this decomposition be applied to solve differential equations?
5. What is the general approach to handling improper fractions before decomposition?

Tip:

When solving for coefficients in partial fractions, systematically organize terms by powers of $x$ to avoid errors!