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Worked solution

Combine a linear factor with an irreducible quadratic

Express 2x2+x−3(x+1)(2x2+3x+4)\dfrac{2x^2+x-3}{(x+1)(2x^2+3x+4)} in partial fractions.

CJC · O Level revision · Example 14

Answer

−23(x+1)+10x−13(2x2+3x+4)-\dfrac2{3(x+1)}+\dfrac{10x-1}{3(2x^2+3x+4)}, for real x≠−1x\ne-1.

Compare your reasoning

Allow a linear numerator over the quadratic

2x2+x−3(x+1)(2x2+3x+4)=Ax+1+Bx+C2x2+3x+4\begin{aligned}&\frac{2x^2+x-3}{(x+1)(2x^2+3x+4)}\\ & =\frac A{x+1}+\frac{Bx+C}{2x^2+3x+4}\end{aligned}

The quadratic has discriminant 32−4(2)(4)=−23<03^2-4(2)(4)=-23<0, so it has no real linear factors. The numerator over it must have degree less than 2. This irreducibility/domain check is independently supplied.

Find A at the linear-factor zero

A=2x2+x−32x2+3x+4∣x=−1=2(−1)2+(−1)−32(−1)2+3(−1)+4=−23A=\left.\frac{2x^2+x-3}{2x^2+3x+4}\right|_{x=-1}=\frac{2(-1)^2+(-1)-3}{2(-1)^2+3(-1)+4}=-\frac23

The source uses cover-up for this coefficient. Its callout says substitution also suffices: independently, clearing denominators then setting x=−1 gives −2=3A. No prior cover-up knowledge is required.

Combine the fractions and compare coefficients

2x2+x−3=A(2x2+3x+4)+(Bx+C)(x+1)=−23(2x2+3x+4)+Bx(x+1)+C(x+1),2=−43+B ⟹ B=2+43=103,−3=−83+C ⟹ C=83−3=−13.\begin{aligned}2x^2+x-3&=A(2x^2+3x+4)+(Bx+C)(x+1)\\&=-\frac23(2x^2+3x+4)+Bx(x+1)+C(x+1),\\2&=-\frac43+B\ \Longrightarrow\ B=2+\frac43=\frac{10}3,\\-3&=-\frac83+C\ \Longrightarrow\ C=\frac83-3=-\frac13.\end{aligned}

Retain the source numerator identity before comparing the x² coefficient and constant.

How did this question go?

Common mistakes

The direct cover-up rule isolates simple linear-factor coefficients and the highest power of a repeated linear factor. It does not directly supply every lower-power coefficient or a quadratic numerator.

Solution notes

Substitution or coefficient comparison is sufficient throughout. The source expressly treats cover-up as optional prior school knowledge.