Exercise 12.3

1. Carry out the following divisions:

(i) 48a3 by 6a

Solution:

48a36a = 486a3a = 8a26a2

(ii) 14x3 by 42x2

Solution:

14x342x2 = 1442x3x2

= x =

(iii) 72a3b4c5 by 8ab2c3

Solution:

72a3b4c58ab2c3 = (728)(a³/a)(b⁴/b²)(c⁵/c³)

= 9a2b2c29a3b2c3

(iv) 11xy2z3 by 55xyz

Solution:

11xy2z355xyz

= (1155)(x/x)(y2y)(z3z)

= ()yz2 = yz25yz5

(v) −54l4m3n2 by 9l2m2n2

Solution:

−54l4m3n29l2m2n2

= −6l2m−7l2m

2. Divide the given polynomial by the given monomial:

(i) 3x2−2x ÷ x

Solution:

3x2−2xx = 3x2x - 2xx

= -

(ii) (5a3b−7ab3) ÷ ab

Solution:

5a3b−7ab3ab = 5a3bab - 7ab3ab

= a2 - b2

(iii) (25x5 – 15x4) ÷ 5x3

Solution:

25x5−15x45x3 = 25x55x3 - 15x45x3

= x2 -

(iv) (4l5 – 6l4 + 8l3) ÷ 2l2

Solution:

4l5−6l4+8l32l2 = 4l52l2 - 6l42l2 + 8l32l2

= 2l36l2 - 3l24l2 +

(v) 15(a3b2c2 – a2b3c2 + a2b2c3) ÷ 3abc

Solution:

15a3b2c2−a2b3c2+a2b2c33abc = (a2bc−ab2c+abc2)

= 5 (a - b + c)(a + b + c)

(vi) (3p3−9p2q−6pq2) ÷ (–3p)

Solution:

3p3−9p2q−6pq2−3p = −p2p2 + + 2q2−2q2

(vii) 23a2b2c2+43ab2c2 ÷ 12abc

Solution:

23a2b2c2+43ab2c212abc

= () abc + bc

= 4bc3a+2−4bc3a+2

3. Workout the following divisions:

(i) (49x – 63) ÷ 7

Solution:

49x−637 = 77x−97

= -

(ii) 12x(8x – 20) ÷ 4(2x – 5)

Solution:

12x8x−2042x−5 = 12x×42x−542x−5

=

(iii) 11a3b37c−35 ÷ 3a2b2c−5

Solution:

11a3b37c−353a2b2c−5

= ab

(iv) 54lmn(l + m)(m + n)(n + l) ÷ 81mn(l + m)(n + l)

Solution:

54lmnl+mm+nn+l81mnl+mn+l

= () (m+n)

(v) 36(x + 4)(x2 + 7x + 10) ÷ 9(x + 4)

Solution:

36x+4x2+7x+109x+4 = (x2 + 7x + 10)

= 4(+)(x+5)

(vi) a(a + 1)(a + 2)(a + 3) ÷ a(a + 3)

Solution:

aa+1a+2a+3aa+3 = (a + 1)(a + 2)(a + 1)(a - 2)

4. Factorize the expressions and divide them as directed:

(i) (x2 + 7x + 12) ÷ (x + 3)

Solution:

x2+7x+12x+3 = x+3x+4x+3

= +

(ii) (x2 – 8x + 12) ÷ (x – 6)

Solution:

x2−8x+12x−6

= (x - 6)(x-2x+2)/(x - 6)

= x-2x+2

(iii) (p2 + 5p + 4) ÷ (p + 1)

Solution:

p2+5p+4p+1

= (p + 1)(p+4p-4)/(p + 1)

= p+4p-4

(iv) 15aba2−7a+10 ÷ 3ba−2

Solution:

15aba2−7a+103ba−2

= 15ab(a-2)(a-5a+5)/3b(a-2)

= (a-5)(a+5)

(v) 15lm(2p2 – 2q2) ÷ 3l(p + q)

Solution:

15lm(2p2 – 2q2)/3l(p + q) = 15lm × 2(2p2 – 2q2)/3l(p + q) = p−qp+qp+qp+q2p+q

= (p-q)

(vi) 26z332z2−18 ÷ 13z24z−3

Solution:

26z332z2−1813z24z−3

= 2z2z2 32z2−184z−3

= (16z2 - 9)

= 2z(4z-34z+3)(4z+3)/(4z-3)

= 2z(4z + 3)3z(4z-3)