Standard +0.3 This is a straightforward improper fraction requiring polynomial long division followed by standard partial fraction decomposition. While it involves an extra step compared to proper fractions, the algebraic manipulation is routine for C4 students and requires no problem-solving insight—just methodical application of a standard technique.
Note: \(\frac{9x^2+20x-10}{(x+2)(3x-1)} \equiv \frac{A}{(x+2)}+\frac{B}{(3x-1)}\) leading to \(A=2, B=-1\) will gain maximum B0B0M1A0.
## Question 3:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A = 3$ (constant term $= 3$) | B1 | Their constant term must equal 3 |
| $9x^2+20x-10 \equiv A(x+2)(3x-1)+B(3x-1)+C(x+2)$ | B1 | Forming a correct identity (can be implied by later working) |
| Attempts to find value of either $B$ or $C$ from identity | M1 | Either substituting values or comparing coefficients |
| $x=-2 \Rightarrow B=2$, $x=\frac{1}{3} \Rightarrow C=-1$ | A1 | Correct values of $B$ and $C$ found using correct identity |
| $\frac{9x^2+20x-10}{(x+2)(3x-1)} \equiv 3 + \frac{2}{(x+2)} - \frac{1}{(3x-1)}$ | | Final answer |
**Note:** $\frac{9x^2+20x-10}{(x+2)(3x-1)} \equiv \frac{A}{(x+2)}+\frac{B}{(3x-1)}$ leading to $A=2, B=-1$ will gain maximum B0B0M1A0.