1. Problem statement: Express the following as partial fractions: $\displaystyle \frac{x^3 - 2x}{x^2 + 3x + 2}$. 2. Strategy and formula: When the numerator degree is greater than or equal to the denominator degree, perform polynomial long division to get a polynomial plus a proper rational function. 3. For proper rational functions with distinct linear factors use the template $$\frac{P(x)}{(x+a)(x+b)}=\frac{A}{x+a}+\frac{B}{x+b}.$$ 4. Polynomial division: Divide $x^3 - 2x$ by $x^2 + 3x + 2$ to separate the polynomial part and the proper fraction. 5. First term of the quotient is x because $x^3/x^2 = x$. 6. Multiply the divisor by x to get $x^3 + 3x^2 + 2x$ and subtract from the numerator to get remainder $-3x^2 -4x$. 7. Next quotient term is -3 because $-3x^2/x^2 = -3$. 8. Multiply the divisor by -3 to get $-3x^2 -9x -6$ and subtract to obtain remainder $5x+6$. 9. Thus $$\frac{x^3 - 2x}{x^2 + 3x + 2} = x - 3 + \frac{5x+6}{x^2+3x+2}. $$ 10. Factor denominator: $x^2+3x+2=(x+1)(x+2)$. 11. Setup partial fractions: Write $$\frac{5x+6}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}. $$ 12. Solve for A and B: Multiply both sides by $(x+1)(x+2)$ to obtain $5x+6 = A(x+2)+B(x+1)$. 13. Expand the right side to get $5x+6 = (A+B)x + (2A+B)$. 14. Equate coefficients: from x-terms $A+B=5$ and from constants $2A+B=6$. 15. Subtract the first equation from the second to get $A=1$. 16. Substitute $A=1$ into $A+B=5$ to get $B=4$. 17. Final answer: Combine the polynomial and partial fractions to get $$\frac{x^3 - 2x}{x^2 + 3x + 2} = x - 3 + \frac{1}{x+1} + \frac{4}{x+2}. $$ 18. This is the partial fraction decomposition.