Moderate -0.3 This is a straightforward partial fractions decomposition with a linear factor and an irreducible quadratic. The setup is standard (A/x + (Bx+C)/(x²+2)) and solving for coefficients requires only routine algebraic manipulation. While it's Further Maths content, it's a textbook exercise with no complications, making it slightly easier than average overall.
For alternative of the form: SC4 allows some alternative algebraic manipulation
$\frac{2x^3 + x + 12}{(2x-1)(x^2+4)} = A + \frac{B}{2x-1} + \frac{Cx+D}{x^2+4}$ | B1 | For correct form soi ($A$ can be $Px+Q$, but not 0)
$2x^3 + x + 12 = A(2x-1)(x^2+4) + B(x^2+4) + (Cx+D)(2x-1)$ | M1 | For multiplying out from their form
$A=1, B=3$ | B1 | For either $A$ or $B$ correct (dep on 1st B1)
$x^3: 2=2A$ | M1 | For equating at least 2 coefficients (or substitute two values for $x$ or one of each)
$x^1: 1=8A-C+2D$ | M1 |
$x^0: 12=-4A+4B-D$ | |
$C=-1, D=-4$ | A1 | A1A1 | For $C, D$ correct
$\Rightarrow 1 + \frac{3}{2x-1} + \frac{-x-4}{x^2+4}$ | A1 | For correct expression www
ALT: Divide out as not proper | B1 | Divide out
$\Rightarrow 1 + \frac{x^2-7x+16}{(2x-1)(x^2+4)}$ | B1 | Writing in this form including 1
$= 1 + \frac{A}{2x-1} + \frac{Bx+C}{x^2+4}$ | |
$x^2: 1=A+2B$ | M1 | For multiplying out from their form
$x: -7=-B+2C$ | M1 | For equating at least 2 coefficients (or substitute two values for $x$ or one of each)
$1: 16=4A-C$ | |
$\Rightarrow A=3, B=-1, C=-4$ | A1 | B correct
$\Rightarrow 1 + \frac{3}{2x-1} + \frac{-x-4}{x^2+4}$ | A1 | For correct expression www
SC $\Rightarrow \frac{3}{2x-1} + \frac{x^2-x}{x^2+4}$ | | For alternative of the form: SC4 allows some alternative algebraic manipulation