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Chapter 12: Factorisation

    Introduction

    What is Factorisation?

    Division of Algebraic Expressions

    Division of Algebraic Expressions Continued (Polynomial ÷ Polynomial)

    Exercise 12.1

    Exercise 12.2

    Exercise 12.3

    What Have We Discussed?

    Enhanced Curriculum Support

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Chapter 12: Factorisation > Exercise 12.3

Exercise 12.3

1. Carry out the following divisions.

Instructions

(i) 28x4÷56x
28x4÷56x = 2×2×7×x×x×x×x2×2×2×7×x =

(ii) −36y3÷9y2

Instructions

−36y3÷9y2

  • Do the division and cancel out the common factors
  • We get:
  • Let's try it out
  • And we have found the answer

Instructions

(iii) 66pq2r3÷11qr2
66pq2r3÷11qr2 = 2×3×11×p×q×q×r×r×r11×q×r×r =

(iv) 34x3y3z3÷51xy2z3

Instructions

34x3y3z3÷51xy2z3

  • Do the division and cancel out the common factors
  • We get:
  • Let's try it out
  • And we have found the answer

(v) 12a8b8÷−6a6b4

Instructions

12a8b8÷−6a6b4

  • Do the division and cancel out the common factors
  • We get:
  • Let's try it out
  • And we have found the answer

2. Divide the given polynomial by the given monomial.

Instructions

(i) 5x2−6x÷3x
5x2−6x÷3x = x5x−63x =

(ii) 3y8−4y6+5y4÷y4

Instructions

3y8−4y6+5y4÷y4

  • Do the division and cancel out the common factors
  • We get:
  • Let's try it out
  • And we have found the answer

Instructions

(iii) 8x3y2z2+x2y3z2+x2y2z3 ÷ 4x2y2z2
8x3y2z2+x2y3z2+x2y2z3 ÷ 4x2y2z2 = 8x2y2z2x+y+z4x2y2z2 =

(iv) x3+2x2+3x ÷ 2x

Instructions

x3+2x2+3x÷2x

  • Do the division and cancel out the common factors
  • We get:
  • Let's try it out
  • And we have found the answer

(v) p3q6−p6q3÷p3q3

Instructions

p3q6−p6q3÷p3q3

  • Do the division and cancel out the common factors
  • We get:
  • Let's try it out
  • And we have found the answer

3. Work out the following divisions.

Instructions

(i) (10x – 25) ÷ 5
(10x–25) ÷ 5 = 52x−55 = +
(ii) (10x – 25) ÷ (2x – 5)
(10x–25) ÷ (2x–5) = 52x−52x−5 =
(iii) 10y(6y + 21) ÷ 5(2y + 7)
10y(6y+21) ÷ 5(2y+7) = 10y×32y+752y+7 =
(iv) 9x2y23z−24 ÷ 27xy(z – 8)
9x2y23z−24 ÷ 27xy(z – 8) = 9x2y2×3z−827xyz−8 =

(v) 96abc(3a – 12)(5b – 30) ÷ 144(a – 4)(b – 6)

Instructions

96abc3a−125b−30÷144a−4b−6

  • Do the division and cancel out the common factors
  • We get:
  • Let's try it out
  • And we have found the answer

4. Divide as directed.

Instructions

(i) 52x+13x+5 ÷ (2x + 1)
52x+13x+52x+1 =
(ii) 26xyx+5y−4 ÷ 13x(y – 4)
26xyx+5y−413xy−4 =

(iii) 52pqr(p + q)(q + r)(r + p) ÷ 104pq(q + r)(r + p)

Instructions

52pqrp+qq+rr+p÷104pqq+rr+p

  • Do the division and cancel out the common factors
  • We get:
  • Let's try it out
  • And we have found the answer

Instructions

(iv) 20y+4y2+5y+3 ÷ 5(y + 4)
20y+4y2+5y+35y+4 = y2+5y+3
(v) x(x + 1)(x + 2)(x + 3) ÷ x(x + 1)
xx+1x+2x+3xx+1 =

5. Factorise the expressions and divide them as directed.

Instructions

(i) y2+7y+10 ÷ (y + 5)
y2+7y+10 = y2+ 2y + 5y + 10 = (y+2)+ (y+2) =
y2+7y+10÷ (y+5) = y+2y+5y+5 =
(ii) m2−14m−32÷m+2
m2−14m−32 = m2+2m−16m−32 = (m+2) – (m+2) =
m2−14m−32÷ (m+2) = m−16m+2m+2 = -

(iii) 5p2−25p+20÷p−1

Instructions

5p2−25p+20÷p−1

  • Factorised form of numerator:
  • Now let's check our answer, taking out the common factor of 5, we get: 5p2−5p+4
  • This can be re-written as: 5p2−5p+4=5p2−p−4p+4=5p−1p−4
  • Now, 5p2−25p+20÷p−1 =
  • Thus, the quotient is 5×p−4 after cancelling the common factor: (p-1)

(iv) 4yzz2+6z−16÷2yz+8

Instructions

Factorise 4yzz2+6z−16

  • Factorised form of numerator:
  • On factorising z2+6z−16, we get: z+8z−2
  • This gives us: 4yzz2+6z−16=4yzz−2z+8
  • Now, 4yzz2+6z−16÷2yz+8 =
  • Thus, the quotient is 2zz−2 after cancelling the common factors: y and z+8

Instructions

(v) 5pqp2−q2 ÷ 2pp+q
5pqp2−q2 ÷ 2pp+q = 5pqp−qp+q2pp+q =

(vi) 12xy9x2−16y2÷4xy3x+4y

Instructions

Factorise 12xy9x2−16y2

  • Factorised form of numerator:
  • Using the identity- a2−b2=a+ba−b, we get: 3x+4y3x−4y
  • This gives us: 12xy9x2−16y2=12xy3x+4y3x−4y
  • Now, 12xy9x2−16y2÷4xy3x+4y =
  • Thus, the quotient is 33x−4y after cancelling the common factors: x, y and 3x+4y

(vii) 39y350y2−98÷26y25y+7

Instructions

Factorise 39y350y2−98

  • Factorised form of numerator:
  • Now let's check our answer, taking out the common factors of 2, 39 and y3, we get: 5y−75y+7
  • This gives us: 39y350y2−98=78×y3×5y+75y−7
  • Now, 39y350y2−98÷26y25y+7 =
  • Thus, the quotient is 3y5y−7 after cancelling the common factors: 26, y2 and 5y+7