Easy -1.2 This is a straightforward partial fractions question requiring factorization of a quadratic denominator into linear factors, then solving for constants using standard methods (cover-up or equating coefficients). It's a routine textbook exercise with no complications like repeated factors or improper fractions, making it easier than average.
## Question 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x^2 - x - 12 = (x-4)(x+3)$ | B1 | Both factors seen |
| $\frac{5x+1}{x^2-x-12} = \frac{A}{x-4} + \frac{B}{x+3}$ | M1 | Setting up partial fractions using their factors. May be implied by correct expression as final answer |
| $5x+1 = A(x+3) + B(x-4)$ | | |
| Substitute $x = -3$ giving $B = 2$ | M1 | Method for finding either $A$ or $B$ |
| Substitute $x = 4$ giving $A = 3$ | A1 | Both $A$ and $B$ correct if clear which denominator they apply to. ISW if an error made only in the transcription to final answer |
| $\frac{5x+1}{x^2-x-12} = \frac{3}{x-4} + \frac{2}{x+3}$ | | |
**Total: [4]**
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