Thank you Joseph O'keefe from Colyton Grammar School for this solution. Andrei Lazanu from Rumania and Laura Hannick, Townley Grammar School for Girls, also sent in good solutions.

Part (1)

$${10x^2-2x+4\over x^3 + x} = {A\over x} + {Bx+C\over x^2+1}$$ In order to work out the constants $A,B,C$, I put the RHS of the identity over a common denominator: $${10x^2-2x+4\over x^3 + x} = {A(x^2+1)+ x(Bx+C)\over x(x^2+1)}$$ Since the denominators are equal, it follows that the numerators of the fraction must also be equal: $$\eqalign{ 10x^2-2x+4 &= A(x^2+1)+ x(Bx+C)\cr &= Ax^2+Bx^2+Cx+A \cr &= (A+B)x^2+Cx+A}$$ By comparing coefficients, it follows that $A+B=10$, $C=-2$ and $A=4$ so $B=6$.

Part (2)

$${A\over x} + {Bx+C\over x^2+1}={A\over x} + {D\over (x-i)} + {E\over (x+i)}.$$ By taking $A/x$ away and replacing $B$ and $C$ by their respective values, and putting $D$ and $E$ over a common denominator: $${6x-2\over (x^2+1)}= {D(x+i)+E(x-i)\over (x-i)(x+i)}.$$ Again, since the denominators are equal, it follows that the numerators are equal so $6x-2=D(x+i)+E(x-i)$.

By comparing coefficients we have $D+E=6$ and $(D-E)i=-2$. Then $6i-2=D(2i)$ so $D=(6i-2)/2i$. Multiply both the numerator and the denominator by $i$ to get a real denominator: $$D={6i^2-2i\over -2} = {-6-2i\over -2} = 3+i$$ Then $E(-2i)= -6i-2$, so $E=3-i$. So $A=4$, $B=6$, $C=-2$, $D=3+i$ and $E=3-i$ $${10x^2-2x+4\over x^3 + x}= {4\over x} + {6x-2\over x^2+1}={4\over x} + {3+i\over (x-i)} + {3-i\over (x+i)}$$