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Chapter 9 व्यंजकों का गुणनखण्ड Ex 9 a Solutions

Question - 1 : -
दिये गये पदों के उभयनिष्ठ गुणनखण्ड लिखिए4
(i) 7xy, 35 x2y2
(ii) 4m2, 6m2, 8m3
(iii) 3a, 21 ab

Answer - 1 : -

(i) 7xy, 35 x2y2
हलः
7xy के गुणनखण्ड = 7 × x × y
35 x2y2 के गुणनखण्ड = 5 × 7 × x × x × y × y
उभय निष्ठ गुणनखण्ड = 7 × x × y
= 7xy

(ii) 4m2, 6m2, 8m3
हलः
4m2 के गुणनखण्ड = 2 × 2 × m × mx
6m2 के गुणनखण्ड = 2 × 3 × m × m
8m3 के गुणनखण्ड = 2 × 2 × 2 × m × m ×m
उभयनिष्ठ गुणनखण्ड = 2 × m × m = 2m2

(iii) 3a, 21 ab
हलः
3a के गुणनखण्ड = 3 × a
21 a के गुणनखण्ड = 3 × 7 × a × b
उभयनिष्ठ गुणनखण्ड = 3 × a = 3a

Question - 2 : -
व्यंजक 7pq + 8qr + 3qs का एक उभयनिष्ठ गुणनखण्ड होगाः
(i) q
(ii)r
(iii) p+q+r
(iv) 3s

Answer - 2 : -

7pq के गुणनखण्ड =7 × q × 1
8qr के गुणनखण्ड =2 × 2 × 2 × q × r
3qs के गुणनखण्ड = 3 × q × s
उभयनिष्ठ गुणनखण्ड = q

Question - 3 : -
व्यंजक b (6a – b) + 2c (6a + b) के पदों का उभयनिष्ठ गुणनखण्ड होगा
(i) b
(ii) 2c
(iii) (ba – b)
(iv) a – b

Answer - 3 : -

b (6a – b) के गुणनखण्ड = b × (6a – b)
2c (6a – b) के गुणनखण्ड =2 × cx (6a – b)
उभयनिष्ठ गुणनखण्ड = (6a – b)

Question - 4 : - निम्नलिखित के गुणनखण्ड कीजिए
(i) 5x_2 – 25xy
(ii) 9a_2 – 6ax
(iii) 36ab – 60 a°be
(iv) 6P+ 8P_2-4P_3
_(v) 3a_2be + 6ab_2c + 9abc_2

Answer - 4 : -

(i) 5x2 –25xy
हल :
5x2 – 25xy
= 5x (x – 5y)

(ii) 9a2 –6ax
हल :
9a2 – 6ax
= 3a (3a – 2x)

(iii) 36ab – 60 a°be
हल :
36 a2b – 60 a2bc
= 12ab(3a – 5ac)

(iv) 6P+ 8P2-4P3
हल :6P +8P2 – 4P3
= 2P (3 + 4P – 2P2)

(v) 3a2be + 6ab2c +9abc2
हलः
3a2bc + 6ab2c +9abc2
= 3abc (a + 2b = 3c)

 

Question - 5 : - निम्नलिखित के गुणनखण्ड कीजिए
(i) x(x – 2) +3 (x – 2)
(ii) 7(a – a) + 7(4 – a)
(iii) 2y (y + 5)-3(y + 5)
(iv) (d-7) +7 (d – 7)
(v) a(a – 5) + 9 (5 – a)
(vi) (z-2) -3 (z-2)
(vii) 17 (a +3) +17 (3 – a)

Answer - 5 : -

(i) x(x –2) +3 (x – 2)
हल :
x(x – 2) + 3 (x – 2)
= (x – 2) (x + 3)

(ii) 7(a – a) + 7(4 – a)
हल :
7(a – a) + 7 (4 – a)
= (a – 4) – 7(a – 4) = 0

(iii) 2y (y + 5)-3(y + 5)
हल :
2y(y + 5) – 3(y + 5)
= (y + 5) (2y – 3)

(iv) (d-7) +7 (d – 7)
हल :(d –7)2  + 7(d – 7)
= (d – y)2 + 7(d – 7)
= (d-y) + (d – 7 + 7)
= (d – 7) × d
=d(d – 7)

(v) a(a – 5) + 9 (5 – a)
हल :
a(a – 5) + 9 (5 – a)
= a(a – b) + 9(5 – a)
= (a – 5) (a – 9)

(vi) (z-2) -3 (z-2)
हल :
(z – 2)2 – 3(z – 2)
= (z – 2) (z – 2 – 3)
= (z – 2) (z – 5)

(vii) 17 (a +3) +17 (3 – a)
हल :
17 (a + 3) + 17 (3 – a)
= 17 (a + 3) – 17 (a – 3)
= 17 (a + 3 – a + 3)
= 17 × 6 = 102

 

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