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Worked solution

Include both powers of a repeated linear factor

Express x2+6x+9(x−2)2(x+1)\dfrac{x^2+6x+9}{(x-2)^2(x+1)} in partial fractions.

CJC · O Level revision · Example 13

Answer

59(x−2)+253(x−2)2+49(x+1)\dfrac5{9(x-2)}+\dfrac{25}{3(x-2)^2}+\dfrac4{9(x+1)}, for x≠2,−1x\ne2,-1.

Compare your reasoning

Include both first and second powers

x2+6x+9(x−2)2(x+1)=Ax−2+B(x−2)2+Cx+1 ⟹ x2+6x+9=A(x−2)(x+1)+B(x+1)+C(x−2)2\begin{aligned}&\frac{x^2+6x+9}{(x-2)^2(x+1)}\\ & =\frac A{x-2}+\frac B{(x-2)^2}\\ &{}+\frac C{x+1}\ \Longrightarrow\ x^2+6x+9\\ & =A(x-2)(x+1)+B(x+1)\\ &{}+C(x-2)^2\end{aligned}

A repeated factor needs a separate term for each power. Omitting the first-power term generally prevents an identity.

Substitute the two factor zeros into the identity

x=2:4+12+9=3B ⟹ B=253,x=−1:1−6+9=9C ⟹ 9C=4 ⟹ C=49.\begin{aligned}x=2:&\quad4+12+9=3B\ \Longrightarrow\ B=\frac{25}3,\\x=-1:&\quad1-6+9=9C\ \Longrightarrow\ 9C=4\ \Longrightarrow\ C=\frac49.\end{aligned}

The printed substitutions first isolate the highest repeated-factor coefficient B, then C.

Use a third convenient value

9=−2A+253+4(49) ⟹ 2A=109 ⟹ A=59\begin{aligned}&9\\ & =-2A+\frac{25}3+4\left(\frac49\right)\ \Longrightarrow\ 2A\\ & =\frac{10}9\ \Longrightarrow\ A\\ & =\frac59\end{aligned}

The remaining A comes from x=0. Preserve the factor 4 multiplying C.

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