HBSE 8th Class Maths Solutions Chapter 14 Factorization Ex 14.1

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 p2q2
(iv) 2x, 3x2, 4
(v) 6abc, 24 ab2, 12a2b
(vi) 16x3, -4x2, 32x
(vii) 10pq, 20qr, 30rp
(viii) 3x2y3, 10x3y2, 6x2y2z
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 p2q2 = 2 × 2 × 7 × p × p × q × q
∴ HCF = 2 × 7 × p × q = 14 pq

HBSE 8th Class Maths Solutions Chapter 14 Factorization Ex 14.1

(iv) 2x = 2 × x
3x2 = 3 × x × x
4 = 2 × 2
∴ HCF = 1

(v) 6abc = 2 × 3 × a × b × c
24 ab2 = 2 × 2 × 2 × 3 × a × b × b
12a2b = 2 × 2 × 3 × a × a × b
∴ HCF = 2 × 3 × a × b
= 6ab

(vi) 16x3 = 2 × 2 × 2 × 2 × x × x × x
-4x2 = (-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

HBSE 8th Class Maths Solutions Chapter 14 Factorization Ex 14.1

(viii) 3x2y3 = 3 × x × x × y × y × y
10 x3 y22 = 2 × 5 × x × x × x y × y
6x2y2z = 2 × 3 × x × x × y × y × z
∴ HCF = x × x × y × y
= x2y2

Question 2.
Factorise the following ex-pressions
(i) 7x – 42
(ii) 6p – 12q
(iii) 7a2 + 14a
(iv) -16z + 20z3
(v) 20l2m + 30alm
(vi) 5x2y – 15xy2
(vii) 10a2 – 15b2 + 20c2
(viii) -4a2 + 4ab – 4ca
(ix) x2yz + xy2z + xyz2
(x) ax2y + bxy2 + cxyz
Solution:
(i) 7a – 42 = 7(x – 6)
(ii) 6p – 12q = 6 (p – 2q)
(iii) 7a2 + 14a = 7a (a + 2)
(iv) -16z + 20z3 = 42 (-4z + 5z2)
= 4z (5z2 – 4z)
(v) 20l2m + 30alm = 10lm (2l + 3a)
(vi) 5x2y – 15xy2 = 5xy (x – 3y)
(vii) 10a2 – 15b2 + 20c2 = 5 (2a2 – 3b2 + 4c2)
(viii) -4a2 + 4ab – 4ca = 4a(-a + b – c)
(ix) x2yz + xy2z + xyz2 = xyz (x + y + z)
(x) ax2y + bxy2 + cxyz = xy (ax + by + cz)

HBSE 8th Class Maths Solutions Chapter 14 Factorization Ex 14.1

Question 3.
Factorise:
(i) x2 + 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) x2 + 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)

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