To decompose the expression $\frac{x^2 + 4x - 6}{(x + 1)(x - 2)}$ into partial fractions, let's go through the following steps:

1. Set up the partial fraction form:
   Since the denominator $(x + 1)(x - 2)$ consists of two distinct linear factors, the partial fraction decomposition will be: $\frac{x^2 + 4x - 6}{(x + 1)(x - 2)} = \frac{A}{x + 1} + \frac{B}{x - 2}$ where $A$ and $B$ are constants to be determined.

2. Clear the denominator:
   Multiply both sides by $(x + 1)(x - 2)$ to eliminate the denominator: $x^2 + 4x - 6 = A(x - 2) + B(x + 1)$

3. Expand the equation:
   Distribute $A$ and $B$ on the right side: $x^2 + 4x - 6 = Ax - 2A + Bx + B$ Combine like terms: $x^2 + 4x - 6 = (A + B)x + (-2A + B)$

4. Set up equations by equating coefficients:
   Match the coefficients of $x^2$, $x$, and the constant term on both sides of the equation:

   • Coefficient of $x$: $A + B = 4$
   • Constant term: $-2A + B = -6$

5. Solve the system of equations: We have the system:

   A + B = 4 \\ -2A + B = -6 \end{cases}$$ Solve for $$A$$ and $$B$$: - From the first equation, $$B = 4 - A$$. - Substitute $$B = 4 - A$$ into the second equation: $$-2A + (4 - A) = -6$$ Simplify: $$-3A + 4 = -6$$ $$-3A = -10$$ $$A = \frac{10}{3}$$ - Substitute $$A = \frac{10}{3}$$ back into $$B = 4 - A$$: $$B = 4 - \frac{10}{3} = \frac{12 - 10}{3} = \frac{2}{3}$$

6. Write the partial fraction decomposition:
   Substitute $A = \frac{10}{3}$ and $B = \frac{2}{3}$ back into the partial fractions: $\frac{x^2 + 4x - 6}{(x + 1)(x - 2)} = \frac{\frac{10}{3}}{x + 1} + \frac{\frac{2}{3}}{x - 2}$ This can also be written as: $\frac{x^2 + 4x - 6}{(x + 1)(x - 2)} = \frac{10}{3(x + 1)} + \frac{2}{3(x - 2)}$

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Would you like further explanation on any part of this process?

Here are some related questions:

1. How would you handle partial fractions with a quadratic factor in the denominator?
2. What if the numerator had a higher degree than the denominator?
3. Can partial fractions be used for integration?
4. How would complex roots affect partial fraction decomposition?
5. How can partial fractions help in solving differential equations?

Tip: When dealing with partial fractions, always check if the numerator degree is lower than the denominator degree. If not, start with polynomial long division.