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, 35x^{2}y^{3}

(iv) 4m^{2}, 6m^{2}, 8m^{3}

(v) 15p, 20qr, 25rp

(vi) 4x^{2}, 6xy, 8y^{2}x

(vii) 12 x^{2}y, 18xy^{2}

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, 35x^{2}y^{3}

7xy = 7 × x × y

35x^{2}y^{3} = 7 × 5 × x × x × y × y × y

∴ Common factors of 7xy, 35x^{2}y^{3}

= 7, x, y, 7x, 7y, xy, 7xy.

iv) 4m^{2}, 6m^{2}, 8m^{3}

4m^{2} = 2 × 2 × m × m

6m^{2} = 2 × 3 × m × m

8m^{3} = 2 × 2 × 2 × m × m × m

∴ Common factors of 4m^{2} , 6m^{2} , 8m^{3}

= 2, m, m^{2}, 2m, 2m^{2}.

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) 4x^{2}, 6xy, 8y^{2}x

4x^{2} = 2 × 2 × x × x

6xy = 2 × 3 × x × y

8y^{2}x = 2 × 2 × 2 × y × y × x

∴ Common factors of 4x^{2}, 6xy, 8xy^{2} = 2, x, 2x.

vii) 12x^{2}y, 18xy^{2}

12x^{2}2y = 2 × 2 × 3 × x × x × y

18xy^{2} = 3 × 3 × 2 × x × y × y

∴ Common factors of 12x^{2}y, 18xy^{2}

= 2,3, x, y, 6, xy, 6x, 6y, 2x, 2y, 3x, 3y, 6xy.

Question 2.

Factorise the following expressions

(i) 5x^{2} – 25xy

(ii) 9a^{2} – 6ax

(iii) 7p^{2} + 49pq

(iv) 36 a^{2}b – 60 a^{2}bc

(v) 3a^{2}bc + 6ab^{2}c + 9abc^{2}

(vi) 4p^{2} + 5pq – 6pq^{2}

(vii) ut + at^{2}

Solution:

(i) 5x^{2} – 25xy

= 5 x × x × – 5 × 5 × x × y

= 5 × x [x – 5 × y]

= 5x [x – 5y]

ii) 9a^{2} – 6ax

= 3 × 3 × a × a – 2 × 3 × a × x

= 3a [3a – 2x]

iii) 7p^{2} + 49pq

= 7 × p × p +7 × 7 × p × q

= 7p[p + 7q]

iv) 36a^{2}b – 60a^{2}bc

= 2 × 2 × 3 × 3 × a × a × b – 2 × 2 × 3 × 5 × a × a × b × c

= 2 × 2 × 3 × a × a × b[3 – 5c]

= 12a^{2}b [3 – 5c]

v) 3a^{2}bc + 6ab^{2}c + 9abc^{2}

= 3 × a × a × b × c + 3 × 2 × a × b × b × c + 3 × 3 × a × b × c × c

= 3abc [a + 2b + 3c]

vi) 4p^{2} + 5pq – 6pq^{2}

= 2 × 2 × p × p + 5 × p × q – 2 × 3 × p × q × q

= p [4p + 5q – 6q^{2}]

vii) ut + at^{2}

= u × t + a × t × t = t [u + at]

Question 3.

Factorise the following:

(i) 3ax – 6xy + 8by – 4bx

(ii) x^{3} + 2x^{2} + 5x + 10

(iii) m^{2} – mn + 4m – 4n

(iv) a^{3} – a^{2}b^{2} – ab + b^{3}

(v) p^{2}q – pr^{2} – pq + r^{2}

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) x^{3} + 2x^{2} + 5x + 10

= (x^{3} + 2x^{2}) + (5x +10)

= (x^{2} × x + 2 × x^{2}) + (5 × x + 5 × 2)

= x^{2}(x + 2) + 5(x + 2)

= (x + 2) (x^{2} + 5)

iii) m^{2} – mn + 4m – 4n

= (m^{2} – mn) + (4m – 4n)

= (m × m – m × n) + (4 × m – 4 × n)

= m(m – n) + 4(m – n)

= (m – n) (m + 4)

iv) a^{3} – a^{2}b^{2} – ab + b^{3}

= (a^{3} – a^{2}b^{2}) – (ab – b^{3})

= (a^{2} × a – a^{2} × b^{2}) – (a × b – b × b^{2})

= a^{2}(a – b^{2}) – b(a – b^{2})

= (a – b^{2}) (a^{2} – b)

v) p^{2}1 – pr^{2} – pq + r^{2}

= (p^{2}q – pr^{2}) – (pq – r^{2})

= (p × p × q – p × r × r) – (pq – r^{2})

= p(pq – r^{2}) – (pq – r^{2}) × 1

= (p – 1) (pq – r^{2})