# AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.1

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.1 Textbook Questions and Answers.

## AP State Syllabus 8th Class Maths Solutions 12th Lesson Factorisation Exercise 12.1 Question 1.
Find the common factors of the given terms in each.

(i) 8x, 24
(ii) 3a, 2lab
(iii) 7xy, 35x2y3
(iv) 4m2, 6m2, 8m3
(v) 15p, 20qr, 25rp
(vi) 4x2, 6xy, 8y2x
(vii) 12 x2y, 18xy2
Solution:
8x = 2 × 2 × 2 × x
24 = 8 × 3 = 2 × 2 × 2 × 3
∴ Common factors of 8x, 24 = 2, 4, 8.

ii) 3a, 2lab
3a = 3 × a
21ab = 7 × 3 × a × b
∴ Common factors of 3a, 21ab = 3, a, 3a.

iii) 7xy, 35x2y3
7xy = 7 × x × y
35x2y3 = 7 × 5 × x × x × y × y × y
∴ Common factors of 7xy, 35x2y3
= 7, x, y, 7x, 7y, xy, 7xy. iv) 4m2, 6m2, 8m3
4m2 = 2 × 2 × m × m
6m2 = 2 × 3 × m × m
8m3 = 2 × 2 × 2 × m × m × m
∴ Common factors of 4m2 , 6m2 , 8m3
= 2, m, m2, 2m, 2m2.

v) 15p, 20qr, 25rp
15p = 3 × 5 × p
20qr = 4 × 5 × q × r
25rp = 5 × 5 × r × p
∴ Common factors of 15p, 20qr, 25rp = 5.

vi) 4x2, 6xy, 8y2x
4x2 = 2 × 2 × x × x
6xy = 2 × 3 × x × y
8y2x = 2 × 2 × 2 × y × y × x
∴ Common factors of 4x2, 6xy, 8xy2 = 2, x, 2x.

vii) 12x2y, 18xy2
12x22y = 2 × 2 × 3 × x × x × y
18xy2 = 3 × 3 × 2 × x × y × y
∴ Common factors of 12x2y, 18xy2
= 2,3, x, y, 6, xy, 6x, 6y, 2x, 2y, 3x, 3y, 6xy. Question 2.
Factorise the following expressions
(i) 5x2 – 25xy
(ii) 9a2 – 6ax
(iii) 7p2 + 49pq
(iv) 36 a2b – 60 a2bc
(v) 3a2bc + 6ab2c + 9abc2
(vi) 4p2 + 5pq – 6pq2
(vii) ut + at2
Solution:
(i) 5x2 – 25xy
= 5 x × x × – 5 × 5 × x × y
= 5 × x [x – 5 × y]
= 5x [x – 5y]

ii) 9a2 – 6ax
= 3 × 3 × a × a – 2 × 3 × a × x
= 3a [3a – 2x]

iii) 7p2 + 49pq
= 7 × p × p +7 × 7 × p × q
= 7p[p + 7q]

iv) 36a2b – 60a2bc
= 2 × 2 × 3 × 3 × a × a × b – 2 × 2 × 3 × 5 × a × a × b × c
= 2 × 2 × 3 × a × a × b[3 – 5c]
= 12a2b [3 – 5c]

v) 3a2bc + 6ab2c + 9abc2
= 3 × a × a × b × c + 3 × 2 × a × b × b × c + 3 × 3 × a × b × c × c
= 3abc [a + 2b + 3c]

vi) 4p2 + 5pq – 6pq2
= 2 × 2 × p × p + 5 × p × q – 2 × 3 × p × q × q
= p [4p + 5q – 6q2] vii) ut + at2
= u × t + a × t × t = t [u + at]

Question 3.
Factorise the following:
(i) 3ax – 6xy + 8by – 4bx
(ii) x3 + 2x2 + 5x + 10
(iii) m2 – mn + 4m – 4n
(iv) a3 – a2b2 – ab + b3
(v) p2q – pr2 – pq + r2
Solution:
i) 3ax – 6xy + 8by – 4ab
= (3ax – 6xy) – (4ab – 8by)
= (3 × a × x – 2 × 3 × x × y)
– (4 ×a × b – 4 × 2 × b × y)
= 3x(a – 2y) – 4b(a – 2y)
= (a – 2y)(3x – 4b)

ii) x3 + 2x2 + 5x + 10
= (x3 + 2x2) + (5x +10)
= (x2 × x + 2 × x2) + (5 × x + 5 × 2)
= x2(x + 2) + 5(x + 2)
= (x + 2) (x2 + 5)

iii) m2 – mn + 4m – 4n
= (m2 – mn) + (4m – 4n)
= (m × m – m × n) + (4 × m – 4 × n)
= m(m – n) + 4(m – n)
= (m – n) (m + 4)

iv) a3 – a2b2 – ab + b3
= (a3 – a2b2) – (ab – b3)
= (a2 × a – a2 × b2) – (a × b – b × b2)
= a2(a – b2) – b(a – b2)
= (a – b2) (a2 – b) v) p21 – pr2 – pq + r2
= (p2q – pr2) – (pq – r2)
= (p × p × q – p × r × r) – (pq – r2)
= p(pq – r2) – (pq – r2) × 1
= (p – 1) (pq – r2)