# AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.3

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.3 Textbook Questions and Answers.

## AP State Syllabus 8th Class Maths Solutions 12th Lesson Factorisation Exercise 12.3

Question 1.
Carry out the following divisions
(i) 48a3 by 6a
(ii) 14x3 by 42x3
(iii) 72a3b4c5 by 8ab2c3
(iv) 11xy2z3 by 55xyz
(v) -54l4m3n2 by 9l2m2n2
Solution:
(i) 48a3 by 6a
48a3 ÷ 6a
= $$\frac{6 \times 8 \times a \times a^{2}}{6 \times a}$$
= 8a2

(ii) 14x3 by 42x3
= 14x3 ÷ 42x3

(iii) 72a3b4c5 by 8ab2c3

(iv) 11xy2z3 by 55xyz
11xy2z3 ÷ 55xyz

(v) -54l4m3n2 by 9l2m2n2
-54l4m3n2 ÷ 9l2m2n2

= -6l2m

Question 2.
Divide the given polynomial by the given monomial
(i) (3x2 – 2x) ÷ x
(ii) (5a3b – 7ab3) ÷ ab
(iii) (25x5 – 15x4) ÷ 5x3
(iv) (4l5 – 6l4 + 8l3) ÷ 2l2
(v) 15 (a3b2c2 – a2b3c2 + a2b2c3 ) ÷ 3abc
(vi) 3p3– 9p2q – 6pq2) ÷ (-3p)
(vii) ($$\frac{2}{3}$$ a2 b2 c2+ $$\frac{4}{3}$$ a b2 c3) ÷ $$\frac{1}{2}$$abc
Solution:
(i) (3x2 – 2x) ÷ x

(ii) (5a3b – 7ab3) ÷ ab

(iii) (25x5 – 15x4) ÷ 5x3

= 5x2 – 3x (or) x(5x – 3)

(iv) (4l5 – 6l4 + 8l3) ÷ 2l2

= 2l2 – 3l2 + 4l = l(2l2 – 3l + 4)

(v) 15 (a3 b2 c2 – a2 b3 c2 + a2 b2 c3 ) ÷ 3abc

= 5[a x abc – b x abc + c x abc ]
= 5abc [a – b + c]

(vi) 3p3– 9p2q – 6pq2) ÷ (-3p)

= -[p2 – 3pq – 2q2]
= 22 + 3pq – p2

(vii) ($$\frac{2}{3}$$ a2b2c2+ $$\frac{4}{3}$$ ab2c3) ÷ $$\frac{1}{2}$$abc

Question 3.
Workout the following divisions:
(i) (49x -63) ÷ 7
(ii) 12x (8x – 20,) ÷ 4(2x – 5)
(iii) 11a3 b3 (7c – 35) ÷ 3a2 b2 (c – 5)
(iv) 54lmn (l + m) (m + n) (n + l) ÷ 8 lmn (l + m) (n +l)
(v) 36(x + 4)(x2 + 7x + 10) ÷ 9(x + 4)
(vi) a(a+1)(a+2)(a + 3) ÷ a(a + 3)
Solution:
(i) (49x -63) ÷ 7

(ii) 12x (8x – 20,) ÷ 4(2x – 5)

(iii) 11a3 b3 (7c – 35) ÷ 3a2 b2 (c – 5)

(iv) 54lmn (l + m) (m + n) (n + l) ÷ 8 lmn (l + m) (n +l)

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

4 ( x2 + 7x + 10)
= 4 ( x2 + 5x + 2x + 10)
= 4 [x( x + 5) +2(x + 5)]
= 4( x + 5) (x + 2)

(vi) a(a+1)(a+2)(a + 3) ÷ a(a + 3)

= ( a + 1)(a + 2)

Question 4.
Factorize the expressions and divide them as directed:
(i) (x2 + 7x + 12) ÷ (x + 3)
(ii) (x2 – 8x + 12) ÷ (x – 6)
(iii) (p2 + 5p + 4,) (p + l)
(iv) 15ab(a2 – 7a + 10) ÷ 3b(a – 2)
(v) 151m (2p2 – 2q2) ÷ 3l(p + q)
(vi) 26z3(32z2 – 18,) ÷ 13z2 (4z – 3)
Solution:
(i) (x2 + 7x + 12) ÷ (x + 3)
(x2 + 7x + 12) ÷ (x + 3)
x2 + 7x + 12 = x2 + 3x + 4x + 12
= x(x + 3) + 4(x + 3)
= (x + 3) (x + 4)

(ii) (x2 – 8x + 12) ÷ (x – 6)
(x2 – 8x + 12) ÷ (x – 6)
x2 – 8x + 12 = x2 – 6x – 2x + 12
= x(x – 6) – 2(x – 6)
= (x – 6) (x – 2)
∴ (x2 – 8x + 12) 4 (x – 6)
= $$\frac{(x-6)(x-2)}{(x-6)}$$ = x – 2

(iii) (p2 + 5p + 4,) (p + 1)
p2 + 5p + 4 = p2 + p + 4p + 4
= p(p + 1) + 4(p + 1)
= (p + 1) (p + 4)
(p2 + 5p + 4) ÷ (p + 1)
= $$\frac{(p+1)(p+4)}{(p+1)}$$ = p + 4

(iv) 15ab(a2 – 7a + 10) ÷ 3b(a – 2)
15ab (a2 – 7a + 10) ÷ 3b (a – 2)
15ab (a2 – 7a + 10) = 15ab (a2 – 5a – 2a + 10)
= 15ab [(a2 – 2a) – (5a -10)]
= 15ab [a(a – 2) – 5(a – 2)]
= 15ab(a – 2)(a – 5)
∴ 15ab (a2 – 7a + 10) ÷ 3b (a – 2)

(v) 151m (2p2 – 2q2) ÷ 3l(p + q)
15lm (2p2 – 2q2) ÷ 3l (p + q)
15lm (2p2 – 2q2) = 15lm x 2(p2 – q2)
= 30lm (p + q) (p – q)
∴ 15lm(2p2 – 2q2) ÷ 3l(p + q)

(vi) 26z3(32z2 – 18,) ÷ 13z2 (4z – 3)
26z3(32z2 – 18) ÷ 13z2 (4z – 3)
26z3(32z2 – 18) = 26z3 (2 x 16z2 – 2 x 9)
= 26z3 x 2 [16z3 – 9]
= 52z3 [(4z)3 – (3)3]
= 52z3 (4z + 3) (4z – 3)
∴ 26z3 (32z2 – 18) ÷ 13z2 (4z – 3)