HBSE 8th Class Maths Solutions Chapter 14 गुणनखंडन Ex 14.1

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

Haryana Board 8th Class Maths Solutions Chapter 14 गुणनखंडन Ex 14.1

प्रश्न 1.
दिये गये पदों में सार्व गुणनखण्ड ज्ञात कीजिए-
(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
हल:
(i) 12x = 2 × 2 × 3 × x
36 = 2 × 2 × 3 × 3
सार्व गुणनखण्ड = 2 × 2 × 3 = 12
∴ सार्व गुणनखण्ड = 12

(ii) 2y, 22xy
2y = 2 × y
22xy = 2 × 11 × x × y
सार्व गुणनखण्ड 2 व y है। तो
सार्व गुणनखण्ड = 2 × y = 2y

(iii) 14pq, 28 p²q²
14pq = 2 × 7 × p × q
28 p²q² = 2 × 2 × 7 × p × p × q × q
∴ सार्व गुणनखण्ड = 2 × 7 × p × q
= 14 pq
∴ सार्व गुणनखण्ड = 14 pq

(iv) 2x, 3x², 4
2x, 3x², 4
2x = 2 × x
3x² = 3 × x × x
4 = 2 × 2 × 1
∴ सार्व गुणनखण्ड = 1

(v) 6abc, 24 ab², 12a²b
6abc = 2 × 3 × a × b × c
24 ab² = 2 × 2 × 2 × 3 × a × b × b
12a²b = 2 × 2 × 3 × a × a × b
सार्व गुणनखण्ड 2, 3, a और b हैं।
∴ सार्व गुणनखण्ड = 2 × 3 × a × b
= 6ab

(vi) 16x³, -4x², 32x
16x³ = 2 × 2 × 2 × 2 × x × x × x
-4x² = -1 × 2 × 2 × x × x
32x = 2 × 2 × 2 × 2 × 2 × x
सार्व गुणनखण्ड 2, 2, और x हैं।
∴ सार्व गुणनखण्ड = 2 × 2 × x = 4x

(vii) 10pq, 20qr, 30rp
10pq = 2 × 5 × p × q
20qr = 2 × 2 × 5 × q × r
30rp = 2 × 3 × 5 × r × p
सार्व गुणनखण्ड 2 और 5 हैं।
∴ सार्व गुणनखण्ड = 2 × 5 = 10

(viii) 3x²y³, 10x³y², 6x²y²z
3x²y³ = 3 × x × x × y × y × y
10 x³ y² = 2 × 5 × x × x × x y × y
6x²y²z = 2 × 3 × x × x × y × y × z
सार्व गुणनखण्ड x, x, y और y हैं।
= x × x × y × y
सार्व गुणनखण्ड = x²y²

प्रश्न 2.
निम्नलिखित व्यंजकों के गुणनखंड कीजिए
(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) 7x – 42
= 7x – 7 × 6
= 7(x – 6)
∴ 7x – 42 = 7(x – 6)

(ii) 6p – 12q
= 6p – 6 × 2q
= 6 (p – 2q)
∴ 6p – 12q = 6 (p – 2q)

(iii) 7a² + 14a
= 7a² + 7 × 2a
= 7a (a + 2)
∴ 7a² + 14a = 7a (a + 2)

(iv) -16z + 20z³
= 42 (-4z + 5z²)
= 4z (5z² – 4z)
∴ -16z + 20z³ = 4z (5z² – 4z)

(v) 20l²m + 30alm
= (2 × 2 × 5 × l × l × m) + (3 × 2 × 5 × a × l × m)
= 2 × 5 lm (2 × l + 3 × a)
= 10lm (2l + 3a)
∴ 20l²m + 30alm = 10lm (2l + 3a)

(vi) 5x²y – 15xy²
= (5 × x × x × y) – (3 × 5 × x × y)
= 5 xy (x – 3y)
∴ 5x²y – 15xy² = 5xy (x – 3y)

(vii) 10a² – 15b² + 20c²
= (2 × 5 × a × a) – (3 × 5 × b × b × b) + (2 × 2 × 5 × c × c)
= 5 × (2 × a × a – 3 × b × b + 4 × c × c)
= 5 (2a² – 3b² + 4c²)
∴ 10a² – 15b² + 20c² = 5 (2a² – 3b² + 4c²)

(viii) -4a² + 4ab – 4ca
= – (4 × a × a) + (4 × a × b) – (4 × c × a)
= 4a(-a + b – c)
∴ -4a² + 4ab – 4ca = 4a(-a + b – c)

(ix) x²yz + xy²z + xyz²
= (x × x × y × z) + (x × y × y × z) + (x × y × z × z)
= xyz (x + y + z)
∴ x²yz + xy²z + xyz² = xyz (x + y + z)

(x) ax²y + bxy² + cxyz
= (a × x × x × y) + (b × x × y × y) + (c × x × y × z)
= xy (ax + by + cz)
∴ ax²y + bxy² + cxyz = xy (ax + by + cz)

प्रश्न 3.
गुणनखंड कीजिए
(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
हल:
(i) x² + xy + 8x + 8y
= (x² + xy) + (8x + 8y)
= x (x + y) + 8 (x + y)
= (x + y)(x + 8)

(ii) 15xy – 6x + 5y – 2
= (15xy – 6x) + (5y – 2)
= 3x (5y – 2) + 1 (5y – 2)
= (5y – 2) (3x + 1)

(iii) ax + bx – ay – by
= (ax + bx) – (ay – by)
= x(a + b) – y (a + b)
= (a + b) (x – y)

(iv) 15pq + 15 + 9q + 25p
= (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) + (7xy – xyz)
= 1 (z – 7) – xy (z – 7)
= (z – 7)(1 – xy)