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Worked solution

Find three distinct linear-factor coefficients by substitution

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

CJC · O Level revision · Example 12

Answer

12(x+1)+56(x−1)−83(2x+1)\dfrac1{2(x+1)}+\dfrac5{6(x-1)}-\dfrac8{3(2x+1)}, for x≠−1,1,−12x\ne-1,1,-\dfrac12.

Compare your reasoning

Assign one constant numerator to each linear factor

2x+3(x+1)(x−1)(2x+1)=Ax+1+Bx−1+C2x+1 ⟹ 2x+3=A(x−1)(2x+1)+B(x+1)(2x+1)+C(x+1)(x−1)\begin{aligned}&\frac{2x+3}{(x+1)(x-1)(2x+1)}\\ & =\frac A{x+1}+\frac B{x-1}+\frac C{2x+1}\ \Longrightarrow\ 2x\\ &{}+3\\ & =A(x-1)(2x+1)\\ &{}+B(x+1)(2x+1)\\ &{}+C(x+1)(x-1)\end{aligned}

The three factors have distinct zeros. Clear the common denominator to obtain a polynomial identity.

Evaluate at the first two factor zeros

x=−1:2(−1)+3=A(−2)(−1) ⟹ A=12,x=1:2+3=B(2)(3) ⟹ 6B=5 ⟹ B=56.\begin{aligned}x=-1:&\quad2(-1)+3=A(-2)(-1)\ \Longrightarrow\ A=\frac12,\\x=1:&\quad2+3=B(2)(3)\ \Longrightarrow\ 6B=5\ \Longrightarrow\ B=\frac56.\end{aligned}

These substitutions are into the cleared polynomial identity, not the undefined original rational function. They isolate A and B, as in the printed working.

Evaluate at the remaining zero

x=−12:−1+3=C(12)(−32) ⟹ −34C=2 ⟹ C=−83x=-\frac12:\quad-1+3=C\left(\frac12\right)\left(-\frac32\right)\ \Longrightarrow\ -\frac34C=2\ \Longrightarrow\ C=-\frac83

At x=−1/2 the first two terms vanish. The negative coefficient C gives the final subtraction.

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