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Toolkit 43

Partial Fraction Decomposition

5x+1(x−1)(x+2)=Ax−1+Bx+2\frac{5x+1}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2}
Partial Fraction Decomposition rewrites a rational expression as a sum of simpler fractions.\text{Partial Fraction Decomposition rewrites a rational expression as a sum of simpler fractions.}
Two Ways to Calculate the Numerators\text{Two Ways to Calculate the Numerators}
Approach 1: Comparing Coefficients\text{Approach 1: Comparing Coefficients}
5x+1(x−1)(x+2)=Ax−1+Bx+2\frac{5x+1}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2}
Multiply both sides by (x−1)(x+2):\text{Multiply both sides by }(x-1)(x+2):
5x+1=A(x+2)+B(x−1)5x+1=A(x+2)+B(x-1)
Expand:\text{Expand:}
5x+1=(A+B)x+(2A−B)5x+1=(A+B)x+(2A-B)
Compare coefficients:\text{Compare coefficients:}
A+B=5A+B=5
2A−B=12A-B=1
Therefore,\text{Therefore,}
A=2,B=3A=2,\qquad B=3
Approach 2: Heaviside Cover-Up Method\text{Approach 2: Heaviside Cover-Up Method}
f(x)=5x+1(x−1)(x+2)=Ax−1+Bx+2f(x)=\frac{5x+1}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2}
Multiply both sides by (x−1):\text{Multiply both sides by }(x-1):
(x−1)f(x)=5x+1x+2=A+B(x−1)x+2(x-1)f(x)=\frac{5x+1}{x+2}=A+\frac{B(x-1)}{x+2}
Substitute x=1:\text{Substitute }x=1:
A=5(1)+11+2=63=2A=\frac{5(1)+1}{1+2}=\frac{6}{3}=2
Similarly, multiply both sides by (x+2):\text{Similarly, multiply both sides by }(x+2):
(x+2)f(x)=5x+1x−1=A(x+2)x−1+B(x+2)f(x)=\frac{5x+1}{x-1}=\frac{A(x+2)}{x-1}+B
Substitute x=−2:\text{Substitute }x=-2:
B=5(−2)+1−2−1=−9−3=3B=\frac{5(-2)+1}{-2-1}=\frac{-9}{-3}=3
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