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Lesson: Factoring Nonmonic Quadratics

Video

05:20

Sample Question Videos

Worksheet • 25 Questions • 6 Videos

Q1:

Factorise fully 4 π‘₯ βˆ’ 3 2 π‘₯ + 2 8 2 .

  • A 4 ( π‘₯ βˆ’ 1 ) ( π‘₯ βˆ’ 7 )
  • B 4 ( π‘₯ + 1 ) ( π‘₯ + 7 )
  • C 4 ( π‘₯ βˆ’ 1 ) ( π‘₯ + 7 )
  • D ( 4 π‘₯ + 1 ) ( π‘₯ βˆ’ 7 )

Q2:

Factorise fully 6 π‘₯ βˆ’ 1 9 π‘₯ + 1 0 2 .

  • A ( 2 π‘₯ βˆ’ 5 ) ( 3 π‘₯ βˆ’ 2 )
  • B ( 6 π‘₯ βˆ’ 5 ) ( π‘₯ βˆ’ 2 )
  • C ( 2 π‘₯ + 5 ) ( 3 π‘₯ + 2 )
  • D ( 2 π‘₯ βˆ’ 5 ) ( 3 π‘₯ + 2 )
  • E ( 2 π‘₯ + 5 ) ( 3 π‘₯ βˆ’ 2 )

Q3:

Solve the equation 4 𝑑 βˆ’ 3 2 𝑑 + 6 4 = 0 2 by factoring.

  • A 𝑑 = 4
  • B 𝑑 = βˆ’ 4
  • C 𝑑 = βˆ’ 8 or 𝑑 = βˆ’ 2
  • D 𝑑 = 8 or 𝑑 = 2

Q4:

Find the solution set of the equation 1 8 π‘₯ + 1 8 π‘₯ βˆ’ 3 6 = 0 2 in ℝ .

  • A { βˆ’ 2 , 1 }
  • B { 2 , βˆ’ 1 }
  • C { βˆ’ 4 , βˆ’ 9 }
  • D { 4 , βˆ’ 9 }

Q5:

Find the solution set of in .

  • A
  • B
  • C
  • D

Q6:

Given that βˆ’ 1 0 is a root of the equation 2 π‘₯ + 1 3 π‘₯ βˆ’ 7 0 = 0 2 , what is the other root?

  • A 7 2
  • B 2 7
  • C βˆ’ 7 2
  • D βˆ’ 2
  • E2

Q7:

Find the solution set of 2 π‘₯ + 5 π‘₯ βˆ’ 7 = 0 2 in ℝ .

  • A  βˆ’ 7 2 , 1 
  • B  βˆ’ 2 7 , βˆ’ 1 
  • C  7 2 , βˆ’ 1 
  • D { 7 , βˆ’ 1 }
  • E { βˆ’ 7 , 1 }

Q8:

Solve the equation 2 ( π‘₯ + 1 ) + 5 ( π‘₯ + 1 ) = 0 2 .

  • A π‘₯ = βˆ’ 1 , π‘₯ = βˆ’ 7 2
  • B π‘₯ = βˆ’ ο„ž 5 2
  • C π‘₯ = βˆ’ 1 , π‘₯ = βˆ’ 5 2
  • D π‘₯ = ο„ž 5 2
  • E π‘₯ = 1

Q9:

Find the solution set of ( 2 𝑦 + 4 ) + ( 𝑦 + 2 ) = 5 2 2 in ℝ .

  • A { βˆ’ 1 , βˆ’ 3 }
  • B { 1 , 3 }
  • C { 5 , 3 }
  • D { βˆ’ 5 , βˆ’ 3 }

Q10:

Find the solution set of π‘₯ ( π‘₯ + 5 ) 4 βˆ’ π‘₯ ( π‘₯ + 1 ) 8 βˆ’ 3 ( π‘₯ + 4 ) 2 + 1 = 0 in ℝ .

  • A { βˆ’ 5 , 8 }
  • B { 2 , βˆ’ 2 0 }
  • C { 5 , βˆ’ 8 }
  • D { βˆ’ 1 , 4 0 }
  • E { 1 , βˆ’ 4 0 }

Q11:

Find the solution set of βˆ’ 7 ( π‘₯ + 7 ) + 9 ( π‘₯ + 7 ) = 0 2 in ℝ .

  • A  βˆ’ 7 , βˆ’ 4 0 7 
  • B  βˆ’ 4 0 7 
  • C  0 , βˆ’ 4 0 7 
  • D  βˆ’ 7 , 5 8 7 
  • E  7 , βˆ’ 4 0 7 

Q12:

The roots of the equation π‘₯ βˆ’ 1 0 π‘₯ + 1 6 = 0 2 are 𝐿 and 𝑀 , where 𝐿 > 𝑀 . Find, in its simplest form, the quadratic equation whose roots are 𝐿 βˆ’ 7 and 2 𝑀 βˆ’ 6 2 .

  • A π‘₯ βˆ’ 3 π‘₯ + 2 = 0 2
  • B π‘₯ βˆ’ 9 π‘₯ βˆ’ 1 0 = 0 2
  • C π‘₯ + 3 π‘₯ + 2 = 0 2
  • D π‘₯ + π‘₯ βˆ’ 6 = 0 2
  • E π‘₯ + π‘₯ βˆ’ 2 = 0 2

Q13:

Find the solution set of the equation 3 π‘₯ βˆ’ 9 π‘₯ + 6 = 0 2 , giving values to the nearest tenth.

  • A { 2 . 0 , 1 . 0 }
  • B { βˆ’ 2 . 0 , βˆ’ 1 . 0 }
  • C { 4 . 0 , 2 . 0 }
  • D { βˆ’ 4 . 0 , βˆ’ 2 . 0 }

Q14:

Find the solution set of π‘₯ + 9 3 π‘₯ + 3 8 = 1 2 2 in ℝ .

  • A  4 , βˆ’ 5 2 
  • B { βˆ’ 4 , 5 }
  • C  βˆ’ 4 , 5 2 
  • D { 4 , βˆ’ 5 }
  • E  βˆ’ 4 , βˆ’ 5 2 

Q15:

Find the solution set of ( 3 π‘₯ + 6 ) = ( 5 π‘₯ βˆ’ 1 1 ) 2 2 in ℝ .

  • A  1 7 2 , 5 8 
  • B  1 7 2 , βˆ’ 1 7 2 
  • C  1 7 2 
  • D  βˆ’ 1 1 5 , 2 
  • E  1 1 5 , βˆ’ 2 

Q16:

Find the solution set of 1 4 4 π‘₯ = 3 6  in ℝ .

  • A  βˆ’ 1 2 , 1 2 
  • B  1 2 
  • C  0 , 1 2 
  • D  βˆ’ 1 2 

Q17:

Find the solution set of 2 ο€Ή π‘₯ + 3 2  = 7 2 2 in ℝ .

  • A { 2 , βˆ’ 2 }
  • B { 4 , βˆ’ 4 }
  • C { 6 , βˆ’ 6 }
  • D { 6 }
  • E { 2 }

Q18:

Given that 𝑦 + 1 𝑦 = 7 9 2 2 , find 𝑦 + 1 𝑦 .

  • A 9 , βˆ’ 9
  • B81
  • C9
  • D8
  • E 8 , βˆ’ 8

Q19:

Answer the following.

Solve 1 6 π‘₯ βˆ’ 2 4 π‘₯ + 9 = 0 2 .

  • A π‘₯ = βˆ’ 3 4
  • B π‘₯ = 3 4
  • C π‘₯ = 3 4 or βˆ’ 3 4

Deduce from the previous question the solution to 1 6 π‘₯ βˆ’ 2 4 π‘₯ + 9 = 0 2 , using a change of variable.

  • A π‘₯ = βˆ’ 4 3
  • B π‘₯ = 4 3
  • C π‘₯ = 4 3 or βˆ’ 4 3

Q20:

Find the solution set of 5 π‘₯ + 1 2 π‘₯ = βˆ’ 7 2 in ℝ .

  • A  βˆ’ 7 5 , βˆ’ 1 
  • B  βˆ’ 5 7 , βˆ’ 1 
  • C  7 5 , 1 
  • D { 7 , 1 }
  • E { βˆ’ 7 , βˆ’ 1 }

Q21:

Find the solution set of βˆ’ 2 π‘₯ + 2 = 0 4 in ℝ .

  • A { 1 , βˆ’ 1 }
  • B { 1 }
  • C βˆ…
  • D { 1 , 0 }

Q22:

Given that π‘Ž 𝑏 = 𝑏 𝑐 = 2 , find the solution set of the equation π‘Ž π‘₯ βˆ’ 2 𝑏 π‘₯ + 𝑐 = 0 2 .

  • A  1 2 
  • B { 4 }
  • C { 2 }
  • D  1 4 
  • E { 1 }

Q23:

Solve the equation ( 2 π‘₯ βˆ’ 3 ) ( 3 π‘₯ + 4 ) = 0 .

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

Q24:

Solve the equation 4 π‘₯ + 4 0 π‘₯ + 4 0 = βˆ’ 6 0 2 by factoring.

  • A π‘₯ = βˆ’ 5
  • B π‘₯ = 5
  • C π‘₯ = 1 or π‘₯ = 2 5
  • D π‘₯ = βˆ’ 1 or π‘₯ = βˆ’ 2 5

Q25:

Solve the equation 5 π‘₯ βˆ’ 1 4 π‘₯ + 1 0 = 1 5 2 by factoring.

  • A π‘₯ = 7 5
  • B π‘₯ = βˆ’ 7
  • C π‘₯ = βˆ’ 7 5
  • D π‘₯ = 7
  • E π‘₯ = 5 7
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