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Let's break down each part of the problem:

### (a) \( \frac{4x}{x^2 + 6x - 7} \)

The denominator is a quadratic expression, \( x^2 + 6x - 7 \), which can be factored. Factoring this:

\[
x^2 + 6x - 7 = (x - 1)(x + 7)
\]

Since the denominator factors into two distinct linear terms, the partial fraction decomposition will have the following form:

\[
\frac{4x}{(x - 1)(x + 7)} = \frac{A}{x - 1} + \frac{B}{x + 7}
\]

Thus, there are **two undetermined coefficients**: \( A \) and \( B \).

---

### (b) \( \frac{x^2}{x^2 + 2x + 5} \)

The denominator, \( x^2 + 2x + 5 \), cannot be factored into real linear factors because the discriminant (\( \Delta = b^2 - 4ac \)) is negative:

\[
\Delta = 2^2 - 4(1)(5) = 4 - 20 = -16
\]

Therefore, the denominator is an **irreducible quadratic**. For such cases, the partial fraction decomposition takes the form:

\[
\frac{x^2}{x^2 + 2x + 5} = \frac{Ax + B}{x^2 + 2x + 5}
\]

Thus, there are **two undetermined coefficients**: \( A \) and \( B \).

---

If you need further details, feel free to ask!

### Questions:
1. How do you factor quadratics in general?
2. Why are there distinct approaches for linear and irreducible quadratic factors?
3. What does it mean for a quadratic to be irreducible?
4. How do you solve for the undetermined coefficients in partial fraction decomposition?
5. What happens when there is a repeated factor in the denominator?

### Tip:
When performing partial fraction decomposition, always check if the denominator can be factored into linear or irreducible quadratic factors, as this determines the form of the decomposition.

Write out the form of the partial fraction decomposition of the function 4x/(x^2 + 6x - 7) and x^2/(x^2 + 2x + 5).

Learn how to perform partial fraction decomposition for the functions 4x/(x^2 + 6x - 7) and x^2/(x^2 + 2x + 5). The problem covers factoring quadratics, dealing with irreducible quadratic denominators, and identifying undetermined coefficients. Suitable for high school students in Grades 10-12.

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