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Worksheet: Dividing a Polynomial by a Binomial Using Factorization

Q1:

Find the value of π‘˜ that makes the expression π‘₯ βˆ’ π‘˜ π‘₯ + 3 0 2 divisible by π‘₯ βˆ’ 5 .

Q2:

Find the value of π‘˜ that makes the expression π‘₯ βˆ’ π‘˜ π‘₯ + 2 0 2 divisible by π‘₯ βˆ’ 5 .

Q3:

Find the value of π‘˜ that makes the expression π‘₯ βˆ’ π‘˜ π‘₯ βˆ’ 1 8 2 divisible by π‘₯ + 2 .

Q4:

The area of a triangle is ο€Ή 1 2 π‘₯ + 3 8 π‘₯ + 2 8  2 cm2 and its base is ( 2 π‘₯ + 4 ) cm. Write an expression for its height.

  • A ( 6 π‘₯ + 1 4 ) cm
  • B ( 6 π‘₯ + 7 ) cm
  • C ( 1 2 π‘₯ + 7 ) cm
  • D ( 1 2 π‘₯ + 1 4 ) cm

Q5:

The area of a triangle is ο€Ή 1 8 π‘₯ + 7 3 π‘₯ + 3 5  2 cm2 and its base is ( 2 π‘₯ + 7 ) cm. Write an expression for its height.

  • A ( 9 π‘₯ + 1 0 ) cm
  • B ( 9 π‘₯ + 5 ) cm
  • C ( 1 8 π‘₯ + 5 ) cm
  • D ( 1 8 π‘₯ + 1 0 ) cm

Q6:

The area of a triangle is ο€Ή 4 8 π‘₯ + 6 2 π‘₯ + 9  2 cm2 and its base is ( 8 π‘₯ + 9 ) cm. Write an expression for its height.

  • A ( 6 π‘₯ + 2 ) cm
  • B ( 6 π‘₯ + 1 ) cm
  • C ( 1 2 π‘₯ + 1 ) cm
  • D ( 1 2 π‘₯ + 2 ) cm

Q7:

Knowing that the length of a rectangle is π‘₯ + 5 and its area is 2 π‘₯ + 9 π‘₯ βˆ’ 5 2 , express the width of the rectangle algebraically.

  • A π‘₯ βˆ’ 1
  • B 2 π‘₯ + 1
  • C π‘₯ + 1
  • D 2 π‘₯ βˆ’ 1
  • E 2 π‘₯ βˆ’ 2

Q8:

What is the width of a rectangle whose area is ο€Ή 4 π‘₯ βˆ’ 3 2 π‘₯ + 6 π‘₯  3 2 4 cm2 and whose length is ο€Ή 8 π‘₯ + 3 π‘₯  2 cm?

  • A 2 π‘₯ 2 cm
  • B ο€Ή 4 π‘₯ βˆ’ 4 π‘₯  2 cm
  • C ο€Ή 2 π‘₯ + 4 π‘₯  2 cm
  • D ο€Ή 2 π‘₯ βˆ’ 4 π‘₯  2 cm

Q9:

What is the width of a rectangle whose area is ο€Ή 1 4 π‘₯ + 7 2 π‘₯ βˆ’ 8 π‘₯  4 2 6 cm2 and whose length is ο€Ή 8 π‘₯ βˆ’ 2 π‘₯  3 cm?

  • A 4 π‘₯ 3 cm
  • B ο€Ή 1 4 π‘₯ + 9 π‘₯  3 cm
  • C ο€Ή 4 π‘₯ βˆ’ 9 π‘₯  3 cm
  • D ο€Ή 4 π‘₯ + 9 π‘₯  3 cm

Q10:

What is the width of a rectangle whose area is ο€Ή 5 9 π‘₯ βˆ’ 3 6 π‘₯ + 7 π‘₯  5 4 6 cm2 and whose length is ο€Ή 9 π‘₯ + π‘₯  2 3 cm?

  • A 7 π‘₯ 3 cm
  • B ο€Ή 5 9 π‘₯ βˆ’ 4 π‘₯  3 2 cm
  • C ο€Ή 7 π‘₯ + 4 π‘₯  3 2 cm
  • D ο€Ή 7 π‘₯ βˆ’ 4 π‘₯  3 2 cm

Q11:

Knowing that the volume of a box is 1 0 π‘₯ + 3 0 π‘₯ βˆ’ 8 π‘₯ βˆ’ 2 4 3 2 , its length is 2, and its width is π‘₯ + 3 , express the height of the box algebraically.

  • A 4 π‘₯ βˆ’ 5 2
  • B 5 π‘₯ + 4 2
  • C 4 π‘₯ + 5 2
  • D 5 π‘₯ βˆ’ 4 2
  • E 5 π‘₯ βˆ’ 2 2

Q12:

Knowing that the length of a rectangle is 2 π‘₯ + 5 and its area is 4 π‘₯ + 1 0 π‘₯ + 6 π‘₯ + 1 5 3 2 , express the width of the rectangle algebraically.

  • A 3 π‘₯ + 2 2
  • B 2 π‘₯ βˆ’ 3 2
  • C 3 π‘₯ βˆ’ 2 2
  • D 2 π‘₯ + 3 2
  • E π‘₯ + 3 2

Q13:

By factoring, find all the solutions to π‘₯ βˆ’ π‘₯ βˆ’ 1 4 π‘₯ + 2 4 = 0 3 2 , given that ( π‘₯ + 4 ) is a factor of π‘₯ βˆ’ π‘₯ βˆ’ 1 4 π‘₯ + 2 4 3 2 .

  • A π‘₯ = βˆ’ 2 , π‘₯ = βˆ’ 3 , π‘₯ = βˆ’ 4
  • B π‘₯ = βˆ’ 2 , π‘₯ = βˆ’ 3 , π‘₯ = 4
  • C π‘₯ = 2 , π‘₯ = 3 , π‘₯ = 4
  • D π‘₯ = 2 , π‘₯ = 3 , π‘₯ = βˆ’ 4
  • E π‘₯ = 2 , π‘₯ = βˆ’ 3 , π‘₯ = 4

Q14:

A rectangle has an area of ο€Ή 𝑦 + 2 𝑦 + 5 𝑦 + 1 0  3 2 cm2 and a width of ( 𝑦 + 2 ) cm. Find its length in terms of 𝑦 and its perimeter when 𝑦 = 4 .

  • AThe length is ( 𝑦 + 5 ) cm, and the perimeter is 30 cm.
  • BThe length is ο€Ή 𝑦 βˆ’ 5  2 cm, and the perimeter is 34 cm.
  • CThe length is ( 𝑦 + 5 ) cm, and the perimeter is 15 cm.
  • DThe length is ο€Ή 𝑦 + 5  2 cm, and the perimeter is 54 cm.
  • EThe length is ο€Ή 𝑦 + 5  2 cm, and the perimeter is 126 cm.

Q15:

A rectangle has an area of ο€Ή 𝑦 + 1 0 𝑦 + 7 𝑦 + 7 0  3 2 cm2 and a width of ( 𝑦 + 1 0 ) cm. Find its length in terms of 𝑦 and its perimeter when 𝑦 = 2 .

  • AThe length is ( 𝑦 + 7 ) cm, and the perimeter is 42 cm.
  • BThe length is ο€Ή 𝑦 + 7  2 cm, and the perimeter is 46 cm.
  • CThe length is ( 𝑦 + 7 ) cm, and the perimeter is 21 cm.
  • DThe length is ο€Ή 𝑦 + 7  2 cm, and the perimeter is 46 cm.
  • EThe length is ο€Ή 𝑦 + 7  2 cm, and the perimeter is 132 cm.

Q16:

β‹― + 5 0 π‘Ž 𝑏 + 3 8 π‘Ž 𝑏 β‹― = βˆ’ 2 1 π‘Ž 𝑏 + 2 5 𝑏 + 1 9 2 .

  • A 4 0 π‘Ž 𝑏 2 2 , 2 π‘Ž 𝑏
  • B 4 2 π‘Ž 𝑏 2 2 , βˆ’ 2 π‘Ž 𝑏
  • C βˆ’ 4 0 π‘Ž 𝑏 2 2 , 2 π‘Ž 𝑏
  • D βˆ’ 4 2 π‘Ž 𝑏 2 2 , 2 π‘Ž 𝑏