Haryana State Board HBSE 8th Class Maths Solutions Chapter 14 Factorization Ex 14.1 Textbook Exercise Questions and Answers.

## Haryana Board 8th Class Maths Solutions Chapter 14 Factorization Exercise 14.1

Question 1.

Find the common factors of the given terms:

(i) 12x, 36

(ii) 2y, 22xy

(iii) 14pq, 28 p^{2}q^{2}

(iv) 2x, 3x^{2}, 4

(v) 6abc, 24 ab^{2}, 12a^{2}b

(vi) 16x^{3}, -4x^{2}, 32x

(vii) 10pq, 20qr, 30rp

(viii) 3x^{2}y^{3}, 10x^{3}y^{2}, 6x^{2}y^{2}z

Solution:

(i) 12x = 2 × 2 × 3 × x

36 = 2 × 2 × 3 × 3

∴ Highest common factor, HCF = 2 × 2 × 3 = 12

(ii) 2y = 2 × y

22xy = 2 × 11 × x × y

∴ HCF = 2 × y = 2y

(iii) 14pq = 2 × 7 × p x q

28 p^{2}q^{2} = 2 × 2 × 7 × p × p × q × q

∴ HCF = 2 × 7 × p × q = 14 pq

(iv) 2x = 2 × x

3x^{2} = 3 × x × x

4 = 2 × 2

∴ HCF = 1

(v) 6abc = 2 × 3 × a × b × c

24 ab^{2} = 2 × 2 × 2 × 3 × a × b × b

12a^{2}b = 2 × 2 × 3 × a × a × b

∴ HCF = 2 × 3 × a × b

= 6ab

(vi) 16x^{3} = 2 × 2 × 2 × 2 × x × x × x

-4x^{2} = (-2) × 2 × x

32x = 2 × 2 × 2 × 2 × 2 × x

∴ HCF = 2 × 2 × x

= 4x

(vii) 10pq = 2 × 5 × p × q

20qr = 2 × 2 × 5 × q × r

30rp = 2 × 3 × 5 × r × p

∴ HCF = 2 × 5

= 10

(viii) 3x^{2}y^{3} = 3 × x × x × y × y × y

10 x^{3} y2^{2} = 2 × 5 × x × x × x y × y

6x^{2}y^{2}z = 2 × 3 × x × x × y × y × z

∴ HCF = x × x × y × y

= x^{2}y^{2}

Question 2.

Factorise the following ex-pressions

(i) 7x – 42

(ii) 6p – 12q

(iii) 7a^{2} + 14a

(iv) -16z + 20z^{3}

(v) 20l^{2}m + 30alm

(vi) 5x^{2}y – 15xy^{2}

(vii) 10a^{2} – 15b^{2} + 20c^{2}

(viii) -4a^{2} + 4ab – 4ca

(ix) x^{2}yz + xy^{2}z + xyz^{2}

(x) ax^{2}y + bxy^{2} + cxyz

Solution:

(i) 7a – 42 = 7(x – 6)

(ii) 6p – 12q = 6 (p – 2q)

(iii) 7a^{2} + 14a = 7a (a + 2)

(iv) -16z + 20z^{3} = 42 (-4z + 5z^{2})

= 4z (5z^{2} – 4z)

(v) 20l^{2}m + 30alm = 10lm (2l + 3a)

(vi) 5x^{2}y – 15xy^{2} = 5xy (x – 3y)

(vii) 10a^{2} – 15b^{2} + 20c^{2} = 5 (2a^{2} – 3b^{2} + 4c^{2})

(viii) -4a^{2} + 4ab – 4ca = 4a(-a + b – c)

(ix) x^{2}yz + xy^{2}z + xyz^{2} = xyz (x + y + z)

(x) ax^{2}y + bxy^{2} + cxyz = xy (ax + by + cz)

Question 3.

Factorise:

(i) x^{2} + xy + 8x + 8y

(ii) 15xy – 6x + 5y – 2

(iii) ax + bx – ay – by

(iv) 15pq + 15 + 9q + 25p

(v) z – 7 + 7xy – xyz

Solution:

(i) x^{2} + xy + 8x + By

= x (x + y) + 8 (x + y)

= (x + y)(x + 8)

(ii) 15xy – 6x + 5y – 2

= 3x (5y – 2) + 1 (5y – 2)

= (5y – 2) (3x + 1)

(iii) ax + bx – ay – by

= x(a + b) – y (a + b)

= (a + b) (x – y)

(iv) 15pq + 15 + 9q + 25p

= 15pq + 9q + 25p + 15

= 3q + (5p + 3) + 5 (5p + 3)

= (5p + 3) (3q + 5)

(v) z – 7 + 7xy – xyz

= z – 7 – xyz + 7xy

= 1 (z – 7) – xy (z – 7)

= (z – 7)(1 – xy)