Worksheet: Partial Fraction with Linear Factors

In this worksheet, we will practice decomposing rational expressions into partial fractions when the denominator has repeated linear factors.

Q1:

Consider the rational expression 𝑅 = 5 𝑥 3 1 𝑥 + 3 9 ( 𝑥 4 ) ( 𝑥 + 1 ) . The following strategy reveals it as a sum of partial fractions.

What is the value of 𝑅 ( 𝑥 4 ) when 𝑥 = 4 ? Let us call this 𝑎 .

So, 𝑅 = 𝑎 ( 𝑥 4 ) + 𝑆 . What is 𝑆 ? Give the factored and simplified form.

  • A 𝑆 = ( 𝑥 2 ) ( 𝑥 4 ) ( 𝑥 + 1 )
  • B 𝑆 = 5 ( 𝑥 2 ) ( 𝑥 4 ) ( 𝑥 + 1 )
  • C 𝑆 = 5 ( 𝑥 2 ) ( 𝑥 4 ) ( 𝑥 + 1 )
  • D 𝑆 = 5 ( 𝑥 + 2 ) ( 𝑥 4 ) ( 𝑥 + 1 )
  • E 𝑆 = ( 𝑥 2 ) ( 𝑥 4 ) ( 𝑥 + 1 )

Repeat the first step with ( 𝑥 4 ) 𝑆 and ( 𝑥 + 1 ) 𝑆 to find 𝑏 and 𝑐 so that 𝑆 = 𝑏 ( 𝑥 4 ) + 𝑐 ( 𝑥 + 1 ) . What, finally, is the partial fraction decomposition of 𝑅 ?

  • A 3 𝑥 + 1 + 2 𝑥 4 1 ( 𝑥 4 )
  • B 3 𝑥 + 1 + 2 𝑥 4 + 1 ( 𝑥 4 )
  • C 3 𝑥 + 1 + 1 𝑥 4 2 ( 𝑥 4 )
  • D 3 𝑥 + 1 2 𝑥 4 1 ( 𝑥 4 )
  • E 3 𝑥 + 1 1 𝑥 4 + 2 ( 𝑥 4 )

Q2:

Express 𝑥 2 ( 𝑥 + 2 ) ( 𝑥 + 1 ) in partial fractions.

  • A 1 𝑥 + 2 + 2 𝑥 + 1 1 ( 𝑥 + 1 )
  • B 2 𝑥 + 2 1 𝑥 + 1 1 ( 𝑥 + 1 )
  • C 2 𝑥 + 2 + 1 ( 𝑥 + 1 )
  • D 2 𝑥 + 2 1 ( 𝑥 + 1 )
  • E 2 𝑥 + 2 1 𝑥 + 1 1 ( 𝑥 + 1 )

Q3:

Determine the partial fraction decomposition of 𝑥 + 𝑥 + 1 𝑥 ( 𝑥 3 ) ( 𝑥 + 1 ) .

  • A 1 4 ( 𝑥 + 1 ) + 1 1 6 ( 𝑥 + 1 ) 1 3 𝑥 7 1 2 ( 𝑥 3 )
  • B 1 4 ( 𝑥 + 1 ) + 1 1 6 ( 𝑥 + 1 ) + 1 3 𝑥 + 1 3 4 8 ( 𝑥 3 )
  • C 1 4 ( 𝑥 + 1 ) + 1 1 6 ( 𝑥 + 1 ) 1 3 𝑥 + 1 3 4 8 ( 𝑥 3 )
  • D 3 2 ( 𝑥 + 1 ) + 1 1 6 ( 𝑥 + 1 ) 1 3 𝑥 + 1 3 4 8 ( 𝑥 3 )
  • E 3 2 ( 𝑥 + 1 ) + 1 1 6 ( 𝑥 + 1 ) + 1 3 𝑥 7 1 2 ( 𝑥 3 )

Q4:

Find 𝐴 and 𝐵 such that 2 𝑥 ( 𝑥 3 ) = 𝐴 𝑥 3 + 𝐵 ( 𝑥 3 ) .

  • A 𝐴 = 6 , 𝐵 = 2
  • B 𝐴 = 2 , 𝐵 = 6
  • C 𝐴 = 2 , 𝐵 = 6
  • D 𝐴 = 2 , 𝐵 = 6
  • E 𝐴 = 2 , 𝐵 = 6

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