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

(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

(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)

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)