# Factorisation

## Exercise 14.3

Question 1: Carry out the following divisions.

(i) 28x^4 ÷ 56x^4

Answer: 28x^4 ÷ 56x^4
= 1/2 x

(ii) –36y^2 ÷ 9y^2

Answer: -4y

(iii) 66pq^2r^3 ÷ 11qr^2

Answer: 6pqr

(iv) 34x^3y^3z^3 ÷ 51xy^2z^3

Answer: 2/3x^2y

(v) 12a^8b^8 ÷ (– 6a^6b^4)

Answer: -2a^2b^4

Question 2: Divide the given polynomial by the given monomial.

(i) (5x^2 – 6x) ÷ 3x

Answer: (5x)/(3) - 2

(ii) (3y^8 – 4y^6 + 5y^4) ÷ y^4

Answer: 3y^4-4y^2+5

(iii) 8(x^3y^2z^2 + x^2y^3z^2 + x^2y^2z^3) ÷ 4x^2y^2z^2

Answer: 2(x+y+z)

(iv) (x^3 + 2x^2 + 3x) ÷ 2x

Answer: 1/2x^2+2x+3/2

(v) (p^3q^6 – p^6q^3) ÷ p^3q^3

Answer: q^3 - p^3

Question 3: Work out the following divisions.

(i) (10x – 25) ÷ 5

Answer: 2x-5

(ii) (10x – 25) ÷ (2x – 5)

(iii) 10y(6y + 21) ÷ 5(2y + 7)

Answer: 2y \xx 3 = 6y

(iv) 9x^2y^2 (3z – 24) ÷ 27xy(z – 8)

Answer: 1/3xy\xx\3=xy

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

Answer: 2/3abc\xx3xx5=10abc

Question 4: Divide as directed:

(a) 5(2x + 1)(3x + 5) ÷ (2x + 1)

Answer: 5(2x + 1)(3x + 5) ÷ (2x + 1)
= 5(3x + 5) = 15x + 25

(b) 26xy(x + 5) (y – 4) ÷ 13x (y -4)

Answer: 26xy(x + 5) (y – 4) ÷ 13x (y -4)
= 2y(x + 5)(y – 4) ÷(y – 4)
= 2Y (x + 5) = 2xy + 10y

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

Answer: 52pqr(p + q) (q + r)( r + p) ÷ 104 pq(q + r)( r + p)
= r (p + q) (q + r) ( r + p) ÷ 2 (q + r) (r + p)
= r (p + q) ÷ 2

(d) 20 ( y + 4) (y^2 + 5y + 3) ÷ 5 (y + 4)

Answer: 20 ( y + 4) (y^2 + 5y + 3) ÷ 5 (y + 4)
= 4 (y + 4) (y^2 + 5y + 3) ÷ (y + 4)
= 4(y^2 + 5y + 3)

(e) x ( x + 1) ( x + 2) (x + 3) ÷ x( x + 1)

Answer: x ( x + 1) ( x + 2) (x + 3) ÷ x( x + 1)
= (x + 2)(x + 3)

Question 5: Factorise the expressions and divide them as directed

(a) (y^2 + 7y + 10) ÷ (y + 5)

Answer: (y^2 + 7y + 10) ÷ (y + 5)
Here, dividend can be factorised by splitting the middle term as follows:
= (y^2 + 5y + 2y + 10) ÷ (y + 5)
= [y (y + 5) + 2 (y + 5) ] ÷ (y + 5)
= (y + 2) (y + 5) ÷ (y + 5)
= y + 2

(b) (m^2 – 14m – 32) ÷ (m + 2)

Answer: (m^2 – 14m – 32) ÷ (m + 2)
Here, dividend can be factorised by splitting the middle term as follows:
= (m^2 – 16m + 2m – 32) ÷ (m + 2)
= [ m(m – 16) + 2(m – 16) ] ÷ (m + 2)
= (m + 2) (m – 16) ÷ (m +2)
= m – 16

(c) 5p^2 – 25p + 20) ÷ (p – 1)

Answer: 5p^2 – 25p + 20) ÷ (p – 1)
Here, dividend can be factorised by splitting the middle term as follows:
= (5p^2 – 5p – 20p + 20) ÷ (p - 1)
= [5p(p – 1) – 20(p – 1) ] ÷ (p – 1 )
= (5p – 20) (p – 1) ÷ (p – 1)
= 5p – 20

(d) 4yz(z^2 + 6z – 16) ÷ 2y (z + 8)

Answer: 4yz(z^2 + 6z – 16) ÷ 2y (z + 8)
= 2z(z^2 + 8z – 2z – 16) ÷ (z + 8)^2
= 2z [ z(z + 8) – 2(z + 8) ] ÷ (z + 8)
= 2z (z – 2) (z + 8) ÷(z + 8)
= 2z (z – 2)