Worksheet: Dividing Rational Expressions

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In this worksheet, we will practice dividing rational expressions by polynomials or rational expressions involving monomials and/or polynomials.

Q1:

Answer the following questions for the rational expressions 5 ๐‘ฅ โˆ’ 4 5 ๐‘ฅ 1 2 ๐‘ฅ โˆ’ 4 ๐‘ฅ 3 2 and 1 5 ๐‘ฅ โˆ’ 4 5 3 ๐‘ฅ 2 .

Evaluate 5 ๐‘ฅ โˆ’ 4 5 ๐‘ฅ 1 2 ๐‘ฅ โˆ’ 4 ๐‘ฅ 3 2 divided by 1 5 ๐‘ฅ โˆ’ 4 5 3 ๐‘ฅ 2 .

  • A 2 5 ( ๐‘ฅ โˆ’ 3 ) ( ๐‘ฅ + 3 ) 4 ๐‘ฅ ( 3 ๐‘ฅ โˆ’ 1 ) 2 2
  • B ๐‘ฅ ( ๐‘ฅ + 3 ) 3 ๐‘ฅ โˆ’ 1 2
  • C ๐‘ฅ ( ๐‘ฅ + 3 ) 3 ๐‘ฅ โˆ’ 2 2
  • D ๐‘ฅ ( ๐‘ฅ + 3 ) 4 ( 3 ๐‘ฅ โˆ’ 1 ) 2
  • E ๐‘ฅ ( ๐‘ฅ + 3 ) ๐‘ฅ + 1 2

Is the result of 5 ๐‘ฅ โˆ’ 4 5 ๐‘ฅ 1 2 ๐‘ฅ โˆ’ 4 ๐‘ฅ 3 2 divided by 1 5 ๐‘ฅ โˆ’ 4 5 3 ๐‘ฅ 2 a rational expression?

  • A yes
  • B no

Would this be true for any rational expression divided by another rational expression?

  • A no
  • B yes

Q2:

Answer the following questions for the rational expressions 6 ( ๐‘ฅ โˆ’ 2 ) 3 ๐‘ฅ โˆ’ 6 ๐‘ฅ ๏Šจ and 6 ๐‘ฅ โˆ’ 3 2 ๐‘ฅ .

Find the product of 6 ( ๐‘ฅ โˆ’ 2 ) 3 ๐‘ฅ โˆ’ 6 ๐‘ฅ ๏Šจ and 6 ๐‘ฅ โˆ’ 3 2 ๐‘ฅ .

  • A ๐‘ฅ + 3 2 ๐‘ฅ ๏Šจ
  • B 3 ( ๐‘ฅ โˆ’ 1 ) ๐‘ฅ ๏Šจ
  • C 3 ( 3 ๐‘ฅ โˆ’ 1 ) ๐‘ฅ ๏Šจ
  • D 3 ( 2 ๐‘ฅ โˆ’ 1 ) ๐‘ฅ ๏Šจ
  • E 2 ๐‘ฅ + 3 2 ๐‘ฅ

Is the product of 6 ( ๐‘ฅ โˆ’ 2 ) 3 ๐‘ฅ โˆ’ 6 ๐‘ฅ ๏Šจ and 6 ๐‘ฅ โˆ’ 3 2 ๐‘ฅ a rational expression?

  • A yes
  • B no

Would this be true for the product of any two rational expressions?

  • A no
  • B yes

Q3:

Simplify the function ๐‘› ( ๐‘ฅ ) = ๐‘ฅ + 5 ๐‘ฅ ๏Šจ + 9 ๐‘ฅ + 2 0 ร— ๐‘ฅ ๏Šจ + 1 5 ๐‘ฅ + 5 4 7 ๐‘ฅ ๏Šจ + 6 9 ๐‘ฅ + 5 4 , and determine its domain.

  • A ๐‘› ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 6 ( ๐‘ฅ + 4 ) ( 7 ๐‘ฅ + 6 ) , domain = โ„ โˆ’ ๏ฌ โˆ’ 9 , โˆ’ 5 , โˆ’ 4 , โˆ’ 6 7 ๏ธ
  • B ๐‘› ( ๐‘ฅ ) = ๐‘ฅ + 6 ( ๐‘ฅ + 4 ) ( 7 ๐‘ฅ + 6 ) , domain = โ„ โˆ’ ๏ฌ โˆ’ 4 , โˆ’ 6 7 ๏ธ
  • C ๐‘› ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 6 ( ๐‘ฅ + 4 ) ( 7 ๐‘ฅ + 6 ) , domain = โ„ โˆ’ ๏ฌ โˆ’ 4 , โˆ’ 6 7 ๏ธ
  • D ๐‘› ( ๐‘ฅ ) = ๐‘ฅ + 6 ( ๐‘ฅ + 4 ) ( 7 ๐‘ฅ + 6 ) , domain = โ„ โˆ’ ๏ฌ โˆ’ 9 , โˆ’ 5 , โˆ’ 4 , โˆ’ 6 7 ๏ธ
  • E ๐‘› ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 6 ( ๐‘ฅ โˆ’ 4 ) ( 7 ๐‘ฅ โˆ’ 6 ) , domain = โ„ โˆ’ ๏ฌ โˆ’ 9 , โˆ’ 5 , โˆ’ 4 , โˆ’ 6 7 ๏ธ

Q4:

Simplify the function ๐‘› ( ๐‘ฅ ) = ๐‘ฅ ๏Šจ โˆ’ 1 6 2 ๐‘ฅ ๏Šจ + 9 ๐‘ฅ รท 9 ๐‘ฅ ๏Šจ โˆ’ 7 2 ๐‘ฅ + 1 4 4 4 ๐‘ฅ ๏Šจ โˆ’ 8 1 .

  • A ๐‘› ( ๐‘ฅ ) = ( ๐‘ฅ โˆ’ 4 ) ( 2 ๐‘ฅ + 9 ) 9 ๐‘ฅ ( ๐‘ฅ + 4 )
  • B ๐‘› ( ๐‘ฅ ) = ๐‘ฅ + 4 9 ๐‘ฅ ( ๐‘ฅ โˆ’ 4 ) ( 2 ๐‘ฅ โˆ’ 9 )
  • C ๐‘› ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 4 9 ๐‘ฅ ( ๐‘ฅ + 4 ) ( 2 ๐‘ฅ + 9 )
  • D ๐‘› ( ๐‘ฅ ) = ( ๐‘ฅ + 4 ) ( 2 ๐‘ฅ โˆ’ 9 ) 9 ๐‘ฅ ( ๐‘ฅ โˆ’ 4 )
  • E ๐‘› ( ๐‘ฅ ) = ( ๐‘ฅ + 4 ) ( 2 ๐‘ฅ โˆ’ 9 ) ๐‘ฅ ( ๐‘ฅ โˆ’ 4 )

Q5:

Given the function ๐‘› ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 6 ๐‘ฅ โˆ’ 1 5 ๐‘ฅ + 5 4 ร— ๐‘ฅ โˆ’ 3 ๐‘ฅ โˆ’ 2 8 2 ๐‘ฅ โˆ’ 1 5 ๐‘ฅ + 7 2 2 2 , evaluate ๐‘› ( 7 ) , if possible.

  • A โˆ’ 2
  • B โˆ’ 1 2
  • C โˆ’ 1 8 8
  • D undefined

Q6:

Simplify 6 ๐‘ฅ โˆ’ 3 ๐‘ฅ 3 ๐‘ฅ โˆ’ 2 ร— 7 ๐‘ฅ โˆ’ 1 4 2 ๐‘ฅ โˆ’ 1 3 2 .

  • A 3 ๐‘ฅ ( 2 ๐‘ฅ โˆ’ 1 ) 7 ( 3 ๐‘ฅ โˆ’ 2 ) ( ๐‘ฅ โˆ’ 2 ) 2 2
  • B 4 2 ๐‘ฅ โˆ’ 1 0 5 ๐‘ฅ + 4 2 ๐‘ฅ 6 ๐‘ฅ โˆ’ 7 ๐‘ฅ + 2 4 3 2 2
  • C 7 ๐‘ฅ ( ๐‘ฅ โˆ’ 2 ) 3 ๐‘ฅ โˆ’ 2 2
  • D 2 1 ๐‘ฅ ( ๐‘ฅ โˆ’ 2 ) 3 ๐‘ฅ โˆ’ 2 2
  • E 3 ๐‘ฅ ( ๐‘ฅ โˆ’ 2 ) 3 ๐‘ฅ โˆ’ 4 2

Q7:

Simplify 4 ๐‘ฅ โˆ’ 3 ๐‘ฅ 2 ๐‘ฅ โˆ’ 1 โ‹… 2 ๐‘ฅ โˆ’ 5 4 ๐‘ฅ โˆ’ 2 2 .

  • A ( 4 ๐‘ฅ โˆ’ 3 ) ( 2 ๐‘ฅ โˆ’ 5 ) 2 ( 2 ๐‘ฅ โˆ’ 1 ) 2
  • B ๐‘ฅ ( 4 ๐‘ฅ โˆ’ 3 ) ( 2 ๐‘ฅ โˆ’ 5 ) 2 ( 2 ๐‘ฅ โˆ’ 1 )
  • C 8 ๐‘ฅ โˆ’ 2 6 ๐‘ฅ + 1 5 ๐‘ฅ 8 ๐‘ฅ โˆ’ 8 ๐‘ฅ + 2 3 2 2
  • D ๐‘ฅ ( 4 ๐‘ฅ โˆ’ 3 ) ( 2 ๐‘ฅ โˆ’ 5 ) 2 ( 2 ๐‘ฅ โˆ’ 1 ) 2
  • E ๐‘ฅ ( 4 ๐‘ฅ โˆ’ 5 ) ( 2 ๐‘ฅ โˆ’ 3 ) 2 ( 2 ๐‘ฅ โˆ’ 1 )

Q8:

Simplify 1 4 ๐‘ฅ โˆ’ 2 1 ๐‘ฅ 4 ๐‘ฅ โˆ’ 2 0 รท 4 ๐‘ฅ โˆ’ 6 2 ๐‘ฅ โˆ’ 1 2 .

  • A 7 ๐‘ฅ ( 2 ๐‘ฅ โˆ’ 3 ) 2 ( ๐‘ฅ โˆ’ 5 ) ( 2 ๐‘ฅ โˆ’ 1 ) 2
  • B 7 ๐‘ฅ โˆ’ 3 ๐‘ฅ 8 ๐‘ฅ โˆ’ 2 0 2
  • C 1 4 ๐‘ฅ โˆ’ 3 ๐‘ฅ 8 ๐‘ฅ + 4 0 2
  • D 7 ๐‘ฅ ( 2 ๐‘ฅ โˆ’ 1 ) 8 ( ๐‘ฅ โˆ’ 5 )
  • E 7 ๐‘ฅ ( 2 ๐‘ฅ โˆ’ 1 ) 8 ( ๐‘ฅ + 5 )

Q9:

Determine the domain of the function ๐‘› ( ๐‘ฅ ) = 3 ๐‘ฅ โˆ’ 1 5 ๐‘ฅ โˆ’ 6 รท 6 ๐‘ฅ โˆ’ 3 0 4 ๐‘ฅ โˆ’ 2 4 .

  • A โ„ โˆ’ { 5 }
  • B โ„ โˆ’ { 6 }
  • C โ„ โˆ’ { โˆ’ 6 , โˆ’ 5 }
  • D โ„ โˆ’ { 5 , 6 }
  • E โ„

Q10:

Find the volume of a cube whose side length is 4 5 ๐‘ฅ .

  • A 4 5 ๐‘ฅ 3
  • B 6 4 1 2 5 ๐‘ฅ
  • C 6 4 1 2 5
  • D 6 4 1 2 5 ๐‘ฅ 3
  • E 1 6 2 5 ๐‘ฅ 2

Q11:

Simplify the function ๐‘› ( ๐‘ฅ ) = ๐‘ฅ ๏Šจ + 1 6 ๐‘ฅ + 6 4 ๐‘ฅ ๏Šจ + 8 ๐‘ฅ ร— 7 ๐‘ฅ โˆ’ 5 6 6 4 โˆ’ ๐‘ฅ ๏Šจ , and determine its domain.

  • A ๐‘› ( ๐‘ฅ ) = 7 ๐‘ฅ , domain = โ„ โˆ’ { โˆ’ 8 , 0 , 8 }
  • B ๐‘› ( ๐‘ฅ ) = โˆ’ 7 ๐‘ฅ , domain = โ„ โˆ’ { 0 }
  • C ๐‘› ( ๐‘ฅ ) = 7 ๐‘ฅ , domain = โ„ โˆ’ { 0 }
  • D ๐‘› ( ๐‘ฅ ) = โˆ’ 7 ๐‘ฅ , domain = โ„ โˆ’ { โˆ’ 8 , 0 , 8 }
  • E ๐‘› ( ๐‘ฅ ) = โˆ’ 1 7 ๐‘ฅ , domain = โ„ โˆ’ { โˆ’ 8 , 0 , 8 }

Q12:

Simplify the function ๐‘› ( ๐‘ฅ ) = ๐‘ฅ + 3 4 3 2 ๐‘ฅ + 1 4 ๐‘ฅ ร— ๐‘ฅ + 3 ๐‘ฅ โˆ’ 7 ๐‘ฅ + 4 9 3 2 2 , and determine its domain.

  • A ๐‘› ( ๐‘ฅ ) = 2 ๐‘ฅ ๐‘ฅ + 3 , domain = โ„ โˆ’ { โˆ’ 7 , 0 }
  • B ๐‘› ( ๐‘ฅ ) = ๐‘ฅ + 3 2 ๐‘ฅ , domain = โ„ โˆ’ { 0 }
  • C ๐‘› ( ๐‘ฅ ) = 2 ๐‘ฅ ๐‘ฅ + 3 , domain = โ„ โˆ’ { 0 }
  • D ๐‘› ( ๐‘ฅ ) = ๐‘ฅ + 3 2 ๐‘ฅ , domain = โ„ โˆ’ { โˆ’ 7 , 0 }
  • E ๐‘› ( ๐‘ฅ ) = ๐‘ฅ 2 ( ๐‘ฅ + 3 ) , domain = โ„ โˆ’ { โˆ’ 7 , 0 }

Q13:

Simplify the function ๐‘› ( ๐‘ฅ ) = ๐‘ฅ ๏Šจ โˆ’ 1 2 ๐‘ฅ + 3 6 ๐‘ฅ ๏Šฉ โˆ’ 2 1 6 รท 7 ๐‘ฅ โˆ’ 4 2 ๐‘ฅ ๏Šจ + 6 ๐‘ฅ + 3 6 , and determine its domain.

  • A ๐‘› ( ๐‘ฅ ) = 7 , domain = โ„ โˆ’ { 6 }
  • B ๐‘› ( ๐‘ฅ ) = 1 7 , domain = โ„
  • C ๐‘› ( ๐‘ฅ ) = 7 , domain = โ„
  • D ๐‘› ( ๐‘ฅ ) = 1 7 , domain = โ„ โˆ’ { 6 }
  • E ๐‘› ( ๐‘ฅ ) = 1 6 , domain = โ„ โˆ’ { 6 }

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