Question: c) $\frac{x+2}{3x-12} + \frac{2x+1}{4-x}$
1. **State the problem:** Factorize the expression $$\frac{x+2}{3x-12} + \frac{2x+1}{4-x}$$.
2. **Factor denominators:**
- Factor $3x-12$ as $3(x-4)$.
- Note that $4-x$ can be rewritten as $-(x-4)$.
3. **Rewrite the expression:**
$$\frac{x+2}{3(x-4)} + \frac{2x+1}{-(x-4)} = \frac{x+2}{3(x-4)} - \frac{2x+1}{x-4}$$
4. **Find common denominator:** The common denominator is $3(x-4)$.
5. **Rewrite second fraction with common denominator:**
$$- \frac{2x+1}{x-4} = - \frac{3(2x+1)}{3(x-4)} = \frac{-3(2x+1)}{3(x-4)}$$
6. **Combine fractions:**
$$\frac{x+2}{3(x-4)} + \frac{-3(2x+1)}{3(x-4)} = \frac{x+2 - 3(2x+1)}{3(x-4)}$$
7. **Simplify numerator:**
$$x+2 - 3(2x+1) = x + 2 - 6x - 3 = (x - 6x) + (2 - 3) = -5x - 1$$
8. **Final expression:**
$$\frac{-5x - 1}{3(x-4)}$$
9. **Factor out negative sign in numerator:**
$$\frac{-(5x + 1)}{3(x-4)}$$
This is the factorized form of the original expression.
**Answer:** $$\frac{-(5x + 1)}{3(x-4)}$$