The problem asks us to decompose the rational function:

$\frac{2x^2 + 5x - 10}{(x - 1)(x + 2)} = \frac{A}{x - 1} + \frac{B}{x + 2}$

This is called partial fraction decomposition. We need to find the constants $A$ and $B$.

Step 1: Multiply both sides by $(x - 1)(x + 2)$

This eliminates the denominators on both sides:

$2x^2 + 5x - 10 = A(x + 2) + B(x - 1)$

Step 2: Expand the right-hand side

$2x^2 + 5x - 10 = A(x + 2) + B(x - 1)$ $2x^2 + 5x - 10 = A(x) + A(2) + B(x) - B(1)$ $2x^2 + 5x - 10 = A x + 2A + B x - B$ $2x^2 + 5x - 10 = (A + B) x + (2A - B)$

Step 3: Equate coefficients

Now, equate the coefficients of $x$ and the constant terms on both sides:

• Coefficient of $x$: $A + B = 5$
• Constant term: $2A - B = -10$

Step 4: Solve the system of equations

We now have a system of two equations:

1. $A + B = 5$
2. $2A - B = -10$

From the first equation: $B = 5 - A$

Substitute this into the second equation: $2A - (5 - A) = -10$ $2A - 5 + A = -10$ $3A - 5 = -10$ $3A = -5$ $A = -\frac{5}{3}$

Now, substitute $A = -\frac{5}{3}$ into $B = 5 - A$: $B = 5 - \left(-\frac{5}{3}\right)$ $B = 5 + \frac{5}{3} = \frac{15}{3} + \frac{5}{3} = \frac{20}{3}$

Step 5: Conclusion

Thus, the values of the constants are:

• $A = -\frac{5}{3}$
• $B = \frac{20}{3}$

Let me know if you'd like further details or explanations. Here are 5 related questions:

1. How would you solve a similar partial fraction decomposition if there were higher-degree terms in the denominator?
2. What if the degree of the numerator is greater than or equal to the denominator? How does the process change?
3. How can you verify the result of a partial fraction decomposition?
4. What applications of partial fractions can be found in calculus, particularly in integration?
5. How does the method differ for irreducible quadratic factors in the denominator?

Tip: Partial fraction decomposition is especially useful when integrating rational functions.