Balaji Class 9 Maths Solutions Chapter 5 Polynomial and their Factors Ex 5.7

Balaji Class 9 Maths Solutions Chapter 5 Polynomial and their Factors Ex 5.7 बहुपद तथा उनके गुणनखण्ड

प्रश्न 1.
निम्न व्यंजकों के गुणनखण्ड ज्ञात कीजिए
(i) x3 – 8y3
(ii) a3 – 0.216
(iii) 16a4 + 54a
(iv) a6 – 7a3 – 8
हलः
(i) x3 – 8y3 = (x)3 – (2y)3 = (x – 2y)(x2 + 4y2 + 2xy)

(ii) a3 – 0.216 = (a)3 – (0.6)3 = (a – 0.6)(a2 – 0.36 + 0.6 a)

(iii) 16a4 + 54a = 2a(8a3 + 27) = 2a[(2a)3 + (3)3] = 2a[(2a + 3)(4a2 + 9 – 6a)]

(iv) a6 – 7a3 – 8 = a – (8 – 1)a3 – 8 = a6 – 8a3 + a3 – 8 = a (a3 – 8) – 1(a3 – 8)
= (a3 – 8)(a3 + 1) = [(a)3 – (2)3][(a)3 + (1)3]
= (a – 2)(a2 + 4 + 2a)(a + 1)(a + 1 – a)

UP Board Solutions

प्रश्न 2.
निम्न व्यंजकों के गुणनखण्ड ज्ञात कीजिए
(i) x3 + 64
(ii) 1 – 125y3
(iii) \frac{p^{3}}{343} + 8q3
(iv) 7m4n – 7mn4
(v) x7y7 – xy
हलः
(i) x3 + 64 = (x)3 + (4)3 = (x + 4)(x2 + 16 – 4x)

(ii) 1-125y 3 = (1)3 – (5y)3 = (1 – 5y)(1 + 25y2 + 5y)

(iii) \frac{p^{3}}{343}+8 q^{3}=\left(\frac{p}{7}\right)^{3}+(2 q)^{3}=\left(\frac{p}{7}+2 q\right)\left(\frac{p^{2}}{49}+4 q^{2}-\frac{2}{7} p q\right)

(iv) 7m4n – 7mn4 = 7mn(m3 – n3) = 7mn[(m)3– (n)3] = 7mn[(m – n)(m2 + n2 + mn)]

(v) x7y7 – xy = xy(x6y6 – 1) = xy[(x2y2)3 – (1)3].
= y[(x2y2 – 1)(x4y4 + 1 + x2y2]
= xy [(xy)2 – (1)2) (x4y4 – 1 + 2x2y2 xy2)]
= xy (xy + 1)(xy – 1)[(x2y2 + 1)2 – (xy)2]
= xy(xy + 1)(xy – 1)(x2y2 + 1 + xy)(x2y2 + 1 – xy)

प्रश्न 3.
निम्न व्यंजकों के गुणनखण्ड ज्ञात कीजिए
(i) (a + 2b)3 – (a – 2b)3
(ii) a3 + b3 + c(a2 – ab + b2)
(iii) x6 – 7x3 – 8
(iv) x3 – 3x2 + 3x + 7
हल:
(i) (a + 2b)3 – (a – 2b)3 = (a + 2b – a + 2b) [(a + 2b)2 + (a – 2b)2 + (a + 2b) (a – 2b)]
= (4b)[a2 + 4b2 + 4ab + a2 + 4b2 – 4ab + a2 – 4b2]
= (4b)[3a2 + 4b2]

(ii) a3 + b3 + c(a2 – ab + b2) = (a + b)(a + b2 – ab) + c(a2 – ab + b)
= (a2 + b2 – ab)[a + b + c]

(iii) x6 – 7x3 – 8 = x6 – (8 – 1)x3 – 8 (∵ 8 = 1 × 8)
= x6 – 8x3 + x3 – 8 = x3 (x3 – 8) +1(x3 – 8) = (x3 – 8)(x3 + 1)
=[(x)3 – (2)3][(x)3 + (1)3] = (x – 2)(x2 + 4 + 2x)(x + 1)(x2 + 1 – x)

(iv) x3 – 3x2 + 3x + 7
माना f(x) = x3 – 3x2 + 3x + 7
x + 1 = 0
x = -1 रखने पर,
= (-1)3 – 3(-1)2 + 3 × (-1) +7
= -1 – 3 – 3 + 7 = -7 + 7 = 0
∵ (x + 1), f(x) का एक गुणनखण्ड है।
अतः
Balaji Class 9 Maths Solutions Chapter 5 Polynomial and their Factors Ex 5.7
अत: x3 – 3x2 + 3x + 7 के गुणनखण्ड (x + 1)(x2 – 4x + 7) है।

UP Board Solutions

प्रश्न 4.
निम्न व्यंजकों को सरल करके उनके मान ज्ञात कीजिए
Balaji Class 9 Maths Solutions Chapter 5 Polynomial and their Factors Ex 5.7
हलः
Balaji Class 9 Maths Solutions Chapter 5 Polynomial and their Factors Ex 5.7

Ex 5.7 Polynomial and their Factors स्वमूल्यांकन परीक्षण (Self Assessment Test)

निम्न के गुणनखण्ड कीजिए।
प्रश्न 1.
27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p (NCERT)
हलः
(3 p)^{3}+\left(-\frac{1}{6}\right)^{3}+3(3 p)\left(-\frac{1}{6}\right)\left(3 p-\frac{1}{6}\right)=\left(3 p-\frac{1}{6}\right)^{3}

प्रश्न 2.
49a2 + 70ab + 25b2
हलः
(7a)2 + 2 × 7a × 5b + (5b)2
सूत्र [x2 + 2xy + y2 = (x + y)] से
= (7a + 5b)2

प्रश्न 3.
4x2 + y2 + z2 – 4xy -2yz + 4xz
हलः
(2x)2 + (-y)2 + (z)2 + 2. (2x)(-y) + 2(-y) (z) + 2. (2x)(z)
= (2x – y + z)2 = (2x – y + z)(2x – y + z)

UP Board Solutions

प्रश्न 4.
64m3 – 343n3
हलः
64m3 – 343n3 = (4m)3 – (7n)3 = (4m – 7n)(16m2 + 49n2 + 28mn)

प्रश्न 5.
27 – 125a3 – 135a + 225a2    (NCERT)
हलः
27 – 125a3 – 135a + 225a2
= (3)3 – (5a) – 45a(3 – 5a) = (3 – 5a)(9 + 25a2 + 15a) – 45a(3 – 5a)
= (3 – 5a)(9 + 25a2 + 15a – 45a) = (3 – 5a)(9 + 25a2 – 30a)
= (3 – 5a)[(3)2 + (5a)2 – 2 × 3 × 5a]
= (3 – 5a) (3 – 5a)2 = (3 – 5a)3

प्रश्न 6.
8a3 – b3 – 12a2b + 6ab2      (NCERT)
हल:
8a33 – b3 – 12a2b + 6ab2
= (2a)3 – (b)3 – 6ab (2a – b) = (2a – b)(4a2 + b2 + 2ab) – 6ab(2a – b)
= (2a – b) [4a2 + b2 + 2ab – 6ab] = (2a – b)[4a2 + b2 – 4ab]
= (2a – b)[(2a)2 + (b2) – 2 × 2a × b]
= (2a – b)(2a – b)2 = (2a – b)3

प्रश्न 7.
64a3 – 27b3 – 144a2b + 108ab2
हलः
(4a)3 – (3b)3 – 36ab(4a – 3b)
= (4a – 3b)(16a2 + 9b2 + 12ab) – 36ab(4a – 3b)
= (4a – 3b)(16a2 + 9b2 + 12ab – 36ab) = (4a – 3b)(16a2 + 9b2 – 24ab)
= (4a – 3b)[(4a)2 + (3b)2 – 2 × 4a × 3b]
= (4a – 3b)[4a – 3b]2 = (4a – 3b)3

प्रश्न 8.
8a3 + b3 + 12a2b + 6ab2 (NCERT)
हलः
(2a) + (b)3 + 6ab(2a + b)
= (2a + b)(4a2 + b2 – 2ab) + 6ab(2a + b)
= (2a + b)[4a2 + b2 – 2ab + 6ab]
= (2a + b)[4a2 + b2 + 4ab]
= (2a + b)[(2a)2 + (b)2 + 2 × 2a × b]
= (2a + b)(2a + b)2 = (2a + b)3

UP Board Solutions

प्रश्न 9.
27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p
हलः
(3 p)^{3}+\left(-\frac{1}{6}\right)^{3}+3(3 p)\left(-\frac{1}{6}\right)\left(3 p-\frac{1}{6}\right)=\left(3 p-\frac{1}{6}\right)^{3}

प्रश्न 10.
27x3 + y3 + z3 – 9xyz
हलः
(3x)3 + (y)3 + (z)3 – 3(3x)(y)(z)
= (3x + y + z)[(3x)2 + y2 + z2 – 3xy – yz – 3xz]
= (3x + y + z)[9x2 + y2 + z2 – 3xy – yz – 3xz]

प्रश्न 11.
सिद्ध कीजिए कि a2 + b2 + 2(ab + bc + ca) = (a + b)(a + b + 2c)
हलः
L.H.S. = a2 + b2 + 2(ab + bc + ca) = a2 + b2 + 2ab + 2bc + 2ca
= (a + b)2 + 2c(a + b) = (a + b)[a + b + 2c] = R.H.S.

प्रश्न 12.
सिद्ध कीजिए कि 8a3 – b3 – 4ax + 2bx = (2a – b)(4a2 + 2ab + b2 – 2x)
हलः
L.H.S. = 8a3 – b3 – 4ax + 2bx
= (2a)3 – (b)3 – 2x(2a – b) = (2a – b)(4a2 + b2 + 2ab) – 2x(2a – b)
= (2a – b)(4a2 + b2 + 2ab – 2x) = R.H.S.

UP Board Solutions

प्रश्न 13.
सिद्ध कीजिए कि \frac{64}{125} x^{3}-8-\frac{96}{25} x^{2}+\frac{48}{5} x=\left(\frac{4 x}{5}-2\right)^{3}
हलः
Balaji Class 9 Maths Solutions Chapter 5 Polynomial and their Factors Ex 5.7

प्रश्न 14.
सिद्ध कीजिए कि a3x3 – 3a2bx2 + 3ab2x – b3 = (ax – b)3 हलः
L.H.S. = ax3 – b3 – 3a2bx2 + 3ab2x
= (ax)3 – (b)3 – 3abx(ax – b) = (ax – b)(a2x2 + b2 + abx) – 3abx(ax – b)
= (ax – b)(a2x2 + b2 + abx – 3abx) = (ax – b)(a2x2 + b2 – 2abx)
= (ax – b)(ax – b)2 = (ax – 1)3 = R. H.S.

प्रश्न 15.
सिद्ध कीजिए कि a3 + b3 + 1 + 3ab = (a – b + 1)(a3 + b3 + ab – a + b + 1)
हलः
L.H.S. = (a)3 + (-b)3 + (1)3 – 3(a)(-b)(1)
= (a – b + 1)(a2 + b2 +1+ ab + b-a)
[∵ x3 + y2 + z3 – 3xy2 = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)]
∴ a3 – b3 + 1 + 3ab = (a – b + 1)(a2 + b2 + ab + a + b + 1) = R.H.S.

प्रश्न 16.
यदि p = 2 – a, तब सिद्ध कीजिए कि a3 + 6ap + p3 – 8 = 0
हलः
L.H.S. =a3 + 6ap + p – 8
= a3 – 8 + 6ap + p3 = (a – 2)(a2 + 4 + 2a) + p(6a + p2)
p = 2 – a रखने पर
= (a − 2)(a2 + 4 + 2a) +(2 – a)(6a + 4 + a2 – 4a)
= (a − 2)(a2 + 4 + 2a) – (a − 2)(a2 + 2a + 4)
= (a – 2)[a2 + 4 + 2a – a2 – 2a – 4] = 0 = R. H.S.

UP Board Solutions

प्रश्न 17.
यदि x = 2y+6, तब सिद्ध कीजिए कि x3 – 8y3 – 36xy – 216 =0 (NCERT Exemplar) .
हल:
L.H.S. = x3 – 8y3 – 36xy – 216
= (x)3 – (2y)3 – 36xy – 216 = (x – 2y)(x2 +4y2 + 2xy) – 36xy – 216
x = 2y + 6 रखने पर
= [2y + 6 – 2y][(2y + 6)2 + 4y2 + 2(2y + 6)y] – 36(2y + 6)y – 216
= 6[4y2 + 36 + 24y + 4y2 + 4y2 + 12y] – 72y2 – 216y – 216 = 0 = R.H.S.

प्रश्न 18.
यदि x + y = -4, तब सिद्ध कीजिए कि x + y3 – 12xy + 64 = 0 (NCERT Exemplar)
हल:
x3 + y3 – 12xy + 64
= (x + y) (x2 + y2 – xy) – 12xy + 64= (-4)(x2 + y2 – xy) – 12xy + 64
= -4x2 – 4y2 + 4xy – 12xy + 64 = -4x2 – 4y2 – 8xy + 64
= -4x2 – 4(-4 – x)2 – 8x(-4 – x) + 64 [y = -4 – x रखने पर]
= -4x2 – 4(16 + x2 + 8x) + 32x + 8x2 + 64 = 0 = R.H.S.

प्रश्न 19.
यदि a2 + b2 + c2 = 250 और ab+ bc + ca = 3, तब सिद्ध कीजिए कि a + b+ c = ±16
हलः
सूत्र (a + b + c) = a2 + b2 + c2 + 2(ab + bc + ca) .
(a + b + c)2 = 250 + 2 × 3 = 250 + 6 = 256
(a + b + c) = \sqrt{256} = ± 16

प्रश्न 20.
सिद्ध कीजिए कि
Balaji Class 9 Maths Solutions Chapter 5 Polynomial and their Factors Ex 5.7
हलः
माना a = 0.013 तथा b = 0.007
Balaji Class 9 Maths Solutions Chapter 5 Polynomial and their Factors Ex 5.7 6
= a + b = 0.013 + 0.007 = 0.020

UP Board Solutions

प्रश्न 21.
यदि a + b ≠ 0 तब सिद्ध कीजिए कि समीकरण a(x – a) = 2ab – b(x – b) का हल x = a + b होता है|
हलः
∵ a(x – a) = 2ab – b(x – b)
⇒ ax – a2 – 2ab + bx – b2 = 0
x(a + b) = a2 + b2 + 2ab
x(a + b) – (a + b)2 = 0
(a + b){x – (a + b)} = 0
a + b ≠ 0 अत: x – (a + b) = 0
x = (a + b)

प्रश्न 22.
सिद्ध कीजिए कि 2(a2 + b2) = (a + b)2 तब a = b
हलः
∵ 2(a + b2) = (a + b)2
⇒ 2(a2 + b2) = a2 + b2 + 2ab
⇒ 2a2 + 2b2 = a2 + b2 + 2ab
⇒ a2 + b2 – 2ab = 0
⇒ (a – b)2 = 0 ⇒ a = b

UP Board Solutions

प्रश्न 23.
सिद्ध कीजिए कि 1 – x2 + 2xy – y2 = (1 + x – y)(1 – x + y)
हलः
L.H.S. = 1 – x2 + 2xy – y2
= 1 – (x2 – 2xy + y2) = 12 – (x – y)2
= (1 + x – y)(1 – x + y) = R.H.S.

प्रश्न 24.
Balaji Class 9 Maths Solutions Chapter 5 Polynomial and their Factors Ex 5.7
हलः
Balaji Class 9 Maths Solutions Chapter 5 Polynomial and their Factors Ex 5.7

प्रश्न 25.
Balaji Class 9 Maths Solutions Chapter 5 Polynomial and their Factors Ex 5.7
हलः
Balaji Class 9 Maths Solutions Chapter 5 Polynomial and their Factors Ex 5.7

प्रश्न 26.
Balaji Class 9 Maths Solutions Chapter 5 Polynomial and their Factors Ex 5.7
हलः
Balaji Class 9 Maths Solutions Chapter 5 Polynomial and their Factors Ex 5.7

Balaji Publications Mathematics Class 9 Solutions

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