# BSE Odisha 8th Class Maths Solutions Algebra Chapter 4 ଉତ୍ପାଦକୀକରଣ Ex 4(d)

Odisha State Board BSE Odisha 8th Class Maths Solutions Algebra Chapter 4 ଉତ୍ପାଦକୀକରଣ Ex 4(c) Textbook Exercise Questions and Answers.

## BSE Odisha Class 8 Maths Solutions Algebra Chapter 4 ଉତ୍ପାଦକୀକରଣ Ex 4(d)

Question 1.
ସୂତ୍ର ପ୍ରୟୋଗ କରି ଉତ୍ପାଦକ ନିର୍ଣ୍ଣୟ କର ।
(i) 4x² + 4x + 1
(ii) 9b² + 12bc + 4c²
(iii) 16x² + 40ab + 25b²
(iv) 49x² + 112xy + 64y²
(v) a2 + 6a²b² + 9b4
ସମାଧାନ :
a² + 2ab + b² = (a + b)²

(i) 4x² + 4x + 1
= (2x)² + 2.2x.1 + (1)²
= (2x + 1)²
= (2x + 1)(2x + 1)

(ii) 9b² + 12bc + 4c²
= (3b)² + 2.3b.2c + (2c)²
= (3b + 2c)²
= (3b + 2c) (3b + 2c)

(iii) 16a² + 40ab + 25b²
= (4a)² + 2.4a . 5b + (5b)²
= (4a + 5b)²
= (4a + 5b) (4a + 5b)

(iv) 49x² + 112xy + 64y²
= (7x)² + 2.7x.8y + (8y)²
= (7x + 8y)²
= (lx + 8y) (7x + 8y)

(v) a4 + 6a²b² + 9b4
= (a²)² + 2.a².3b² + (3b²)²
= (a² + 3b²)²
= (a² + 3b²)(a² + 3b²)

Question 2.
ସୂତ୍ର ପ୍ରୟୋଗ କରି ଉତ୍ପାଦକ ନିର୍ଣ୍ଣୟ କର ।
(i) 9x² – 6x + 1
(ii) 16x² – 40ab + 25y²
(iii) 49a² – 126ab + 81b²
(iv) 64a² – 16a + 1
(v) 100a2 – 2a²b + b²
ସମାଧାନ :
(1) 9x² – 6x + 1
= (3x)² – 2.3x.1 + (1)²
= (3x – 1)²
= (3x – 1)(3x – 1)

(ii) 16x² – 40xy + 25y²
= (4x)² – 2.4x.5y + (5y)²
= (4x – 5)
= (4x – 5y) (4x – 5y)

(iii) 49a² – 126ab + 81b²
= (7a)² – 2.7a.9b + (9b)²
= (7a – 9b)²
= (7a – 9b)(7a – 9b)

(iv) 64a² – 16a + 1
=(8a)² – 2.8a.1 +(1)²
= (8a – 1)²
= (8a – 1)(8a – 1)

(v) 100a4 – 20a²b + b²
= (10a²)² – 2.10a².b + (b)²
= (10a² – b)²= (10a² – b) (10a² – b)

Question 3.
(i) 16x² + 9y² + 25z² + 24xy + 40xz + 30yz
ସମାଧାନ :
[ a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²]
16x² + 9y² + 25z² + 24xy + 40xz + 30yz
= (4x)² + (3y)² + (5z)² + 2.4x.3y + 2.3y.5z + 2.5z.4x
= (4x + 3y + 5z)² = (4x + 3y + 5z) (4x + 3y + 5z)

(ii) 49x² + 25y² + z² + 70xy + 10yz + 14xz .
ସମାଧାନ :
[ a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²]
49x² + 25y² + z² + 70xy + 10yz + 14xz
= (7x)² + (5y)² + (z)² + 2.7x.5y + 2.5y.z + 2.z.7x
= (7x + 5y + z)² = (7x + 5y + z)(7x + 5y + z)

(iii) 4a² + 9b² + c² + 12ab – 4ac – 6bc
ସମାଧାନ :
[ a² + b² + c² + 2ab – 2bc – 2ca = (a + b – c)² ଅଥବା (c – a – b)²]
4a² + 9b² + c² + 12ab – 4ac – 6bc
= 4a² + 9b² + c² + 12ab – 6bc – 4ac
= (2a)² + (3b)² + (c)² + 2.2a . 3b – 2.3b . c – 2 . c . 2a
= (2a + 3b – c)² = (2a + 3b – c) (2a + 3b – c)
ବିକଳ୍ପ ସମାଧାନ :
4a² + 9b² + c² + 12ab – 4ac – 6bc
= (2a)² + (3b)² + (- c)² + 2.2a.3b + 2.2a (-c) + 2.3b (-c)
ଏଠାରେ 2a = m, 3b = n, – c = p
m² + n² + p² + 2mn + 2mp + 2np = (m + n + p)² = (2a + 3b – c)
4a² + 9b² + c² + 12ab – 4ac – 6bc = (2a + 3b – c)² = (2a + 3b – c) (2a + 3b – c)

(iv) 100a² + 81b² + 49c² – 180ab – 140ac + 126bc
ସମାଧାନ :
[ a² + b² + c² – 2ab + 2bc – 2ca = (a – b – c)²] ଅଥବା (b + c – a)²
100a² + 81b² + 49c² – 180ab – 140ac + 126bc
= 100a² + 81b² + 49c² – 180ab + 126bc – 140ac
= (10a)² + (9b)² + (7c)² – 2 . 10a . 9b + 2.9b . 7c – 2.7c . 10a
= (10a – 9b – 7c)² = (10a – 9b – 7c)(10a – 9b – 7c)

(v) x4 + y² + z² – 2x²y – 2x²z + 2yz
ସମାଧାନ :
[ a² + b² + c² – 2ab + 2bc – 2ca = (a – b – c)²]
x4 + y² + z2 – 2x²y – 2x²z + 2yz
= x4 + y² + z2 – 2x²y + 2yz – 2x²z
= (x²)² + (y)² + (z)² – 2.x².y + 2.y.z-2.z.x² = (x² – y – z)² = (x² – y – z)(x² – y – z)

Question 4.
ସୂତ୍ର ପ୍ରୟୋଗ କରି ଉତ୍ପାଦକ ନିର୍ଣ୍ଣୟ କର ।
(i) 16a² – 9b²
ସମାଧାନ :
a² – b² = (a + b)(a – b)
16a² – 9b² = (4a)² – (3b)²
= (4a + 3b)(4a – 3b)
16a² – 9b² = (4a + 3b)(4a – 3b)

(ii) 25a² – 36b²
ସମାଧାନ :
[a² – b² = (a + b)(a – b)]
25a² – 36b²
= (5a)² – (6b)²
= (5a + 6b)(5a – 6b)
25a² – 36b² = (5a + 6b)(5a – 6b)

(iii) 81a² – 100b²
ସମାଧାନ :
[a² – b² = (a + b)(a – b)]
81a² – 100b²
= (9a)² – (10b)² (ix)
= (9a + 10b)(9a – 10b)
81a² – 100b² = (9a + 10b)(9a – 10b)

(iv) 16a² – 49b²
ସମାଧାନ :
[a² – b² = (a + b)(a – b)]
16a² – 49b²
= (4a)² – (7b)²
= (4a + 7b)(4a – 7b)
16a² – 49b² = (4a + 7b)(4a – 7b)

(v) 144a² – 225b²
ସମାଧାନ :
[a² – b² = (a + b)(a – b)]
144a² – 225b²
= (12a)² – (15b)²
= (12a + 15b)(12a – 15b)
144a² – 225b² = (12a + 15b)(12a – 15b)

(vi) 256a² – 289b²
ସମାଧାନ :
[a² – b² = (a + b)(a – b)]
256a² – 189b²
= (16a)² – (17b)²
= (16a + 17b)(16a – 17b)
256a² – 289b² = (16a + 17b)(16a – 17b)

(vii) 400a² – 225b²
ସମାଧାନ :
[a² – b² = (a + b)(a – b)]
400a² – 225b²
= 25 (16a² – 9b²)
= 25 {(4a)² – (3b)²}
= 25 (4a + 3b)(4a – 3b)
400a² – 225b² = 25 (4a + 3b)(4a – 3b)

(viii) 441a² – 900b²
ସମାଧାନ :
[a² – b² = (a + b)(a – b)]
441a² – 900b² = 9 (49a² – 100b²)
= 9 (7a + 10b)(7a – 10b)
441a² – 900b² = 9 (7a + 10b)(7a – 10b)’

(ix) 121a² – 289b²
ସମାଧାନ :
[a² – b² = (a + b)(a – b)]
121a² – 289b² = (11a)² – (17b)²’
= (11a + 17b)(11a – 17b)
121a² – 289b² = (11a + 17b)(11a – 17b)

(x) 81a² – 361b²
ସମାଧାନ :
[ a² – b² = (a + b)(a – b)]
81a² – 361b² = (9a)² 7 (19b)² = (9a + 19b)(9a – 19b)
81a² – 361b² = (9a + 19b)(9a – 19b)

(xi) (a + b)² – c²
ସମାଧାନ :
[ a² – b² = (a + b)(a – b)]
(a + b)² – c²
= {(a + b) + c} {(a + b) – c}
(a + b)² – c² = (a + b + c)(a + b – c)

(xii) a² – (b – c)²
ସମାଧାନ :
[ a² – b² = (a + b)(a – b)]
a² – (b – c)²
ଏଠାରେ b – c = x ନେଲେ, ଦତ୍ତ ପରିପ୍ରକାଶଟି ହେବ a² – x²
a² – x² = (a + x)(a – x)
= {a + (b – c)} {a – (b – c)}
= (a + b – c)(a – b + c)
a² – (b – c)² = (a + b – c)(a – b + c)

Question 5.
ସୂତ୍ର ପ୍ରୟୋଗ କରି ଉତ୍ପାଦକ ନିର୍ଣ୍ଣୟ କର ।
(i) a4 + a² + 1
(ii) 4x4 + 1
(iii) x4 + 36x²y² + 1296y4
(iv) x4 + 9x²y² + 81y4
(v) x4 + 16x² + 256
ସମାଧାନ :
[a4 +a²b² + b4 =(a² + ab + b²)(a² – ab + b²)]
(i) a4 + a² + 1 = a4 + a².1.1 + 1 = (a)4 + (a)² . (1)² + (1)4
= {a² + a . 1 + (1)²}{a² – a . 1 + (1)²} = (a² + a + 1)(a² – a + 1)
a4 + a² + 1 = (a² + a + 1)(a² – a + 1)
ବିକଳ୍ପ ସମାଧାନ :
a4 + a² + 1 = a4 + 1 + a² = (a²)² + (1)² + 2a² – a² = (a² + 1)² – (a)²
= (a² + 1 + a)(a² + 1 – a) = (a² + a + 1)(a² – a + 1)
a4 + a² + 1 = (a² + a + 1)(a² – a + 1)

(ii) 4x4 + 1 = (2x²)² + (1)² =t (2x²)² + (1)² + 2.2x².1 – 2.2x².1
= (2x² + 1)² – 4x² = (2x² + 1)² – (2x)²
= (2x² + 1 + 2x) (2x² + 1 – 2x) = (2x² + 2x + 1) (2x² – 2x + 1)

(iii) x4 + 36x²y² + 1296y4
= (x)4 + x² . (6y)² + (6y)4
= {x² + x . 6y + (6y)²}{x² – x.6y + (6y)²}
= (x² + 6xy + 36y²)(x² – 6xy + 36y²)
x4 + 36x²y² + 1296y4 = (x² + 6xy + 36y²)(x² – 6xy + 36y²)

(iv) x4 + 9x²y² + 81y4
= (x)4 + (x)² . (3y)² + (3y)4
= {x² + x . 3y + (3y)²}{x² – x . 3y + (3y)²}
= (x² + 3xy + 9y²)(x² – 3xy + 9y²)
x4 + 9x²y² + 81y4 = (x² + 3xy + 9y²)(x² – 3xy + 9y²)

(v) x4 + 16x² + 256
= (x)4 + x².42 + (4)4= (x² + x . 4 + 42)(x² – x . 4 + 42)
= (x² + 4x + 16)(x² – 4x + 16)
x4 + 16x² + 256 = (x² + 4x + 16)(x² – 4x + 16)

Question 6.
ସୂତ୍ର ପ୍ରୟୋଗ କରି ଉତ୍ପାଦକ ନିର୍ଣ୍ଣୟ କର ।
(i) a² + 6a + 9 – b²
(ii) a² – 4a + 4 – c²
(iii) 4a² – 4a + 1 – 9b²
(iv) a² – 6ab + 9b² – 16c²
(v) 16a² – 24ab + 9b² – 25c²
ସମାଧାନ :
(i) a² + 6a + 9 – b²
= {(a)² + 2 . a . 3 + (3)²} – (b)²
= (a + 3)² – (b)² = (a + 3 + b)(a + 3 – b)
= (a + b + 3)(a – b + 3)
a² + 6a + 9 – b² = (a + b + 3)(a – b + 3)

(ii) a² – 4a + 4 – c² = {(a)² – 2 . a . 2 + (2)²} – (c)²
= (a – 2)² – (c)² = (a – 2 + c)(a – 2 – c) = (a + c – 2)(a – c – 2)
a² – 4a + 4 – c² = (a + c – 2)(a – c – 2)

(iii) 4a² – 4a + 1 – 9b² = {(2a)² – 2 . 2a . 1 + (1)²} – (3b)²
= (2a – 1)² – (3b)² = (2a – 1 + 3b)(2a – 1 – 3b) = (2a + 3b – 1)(2a – 3b – 1)
4a² – 4a + 1 – 9b² = (2a + 3b – 1)(2a – 3b – 1)

(iv) a² – 6ab + 9b² – 16c²
= {(a)² – 2 . a . 3b + (3b)²} – (4c)² = (a – 3b)² – (4c)² = (a – 3b + 4c)(a – 3b – 4c)
a² – 6ab + 9b² – 16c² = (a – 3b + 4c)(a – 3b – 4c)

(v) 16a² – 24ab + 9b² – 25c²
= {(4a)² – 2.4a . 3b + (3b)²} – (5c)² = (4a – 3b)² – (5c)²
= (4a – 3b + 5c)(4a – 3b – 5c)
16a² – 24ab + 9b² – 25c² = (4a – 3b + 5c)(4a – 3b – 5c)

Question 7.
ଦୁଇଟି-ପୂର୍ବବର୍ଗର ଅନ୍ତର ରୂପେ ପ୍ରକାଶ କରି ଉତ୍ପାଦକ ନିର୍ଣ୍ଣୟ କର ।
(i) x² – 2x – 195
(ii) x² + 4x – 357
(iii) x² + 6x – 112
(iv) x² + 2x – 899
(v) x² – 4x – 621
(vi) x² – 10x – 171
(vii) x² – 6x – 891
(viii) y² + 4x – 192
ସମାଧାନ :
(i) x² – 2x – 195
= x² – 2.x.1 + 1 – 1 – 195
= {(x)² – 2 . x . 1 + (1)²} – 196
= (x – 1)² – (14)²
= (x – 1 + 14)(x – 1 – 14)
= (x + 13)(x – 15)
x² – 2x – 195 = (x + 13)(x – 15)

(ii) x² + 4x – 357
= x² + 4x + 4 – 4 – 357
= (x)² + 2.x.2 + (2)² – 361
= (x + 2)² – (19)²
= (x + 2 + 19)(x + 2-19)
= (x + 21) (x – 17)
x² + 4x – 357 = (x + 21)(x – 17)

(iii) x² + 6x-112
= x² + 6x + 9 – 9 – 112
= {(x)² + 2 . x . 3 + (3)²} – 121
= (x + 3)² – (11)²
= (x + 3 + 11)(x + 3 – 11)
= (x + 14) (x – 8)
x² + 6x – 112 = (x + 14)(x – 8)

(iv) x² + 2x – 899
= x² + 2x + 1 – 1 – 899
= {x² + 2 . x . 1 + (l)²} – 900
= (x + l)² – (30)²
= (x + 1 + 30)(x + 1 – 30)
= (x + 31)(x – 29)
x² + 2x – 899 = (x + 31)(x – 29)

(v) x² – 4x – 621
= x² – 4x + 4 – 4 – 621
= {x² – 2 . x . 2 + (2)²} – 625
= (x – 2)² – (25)²
= (x – 2 + 25)(x – 2 – 25)
= (x + 23) (x – 27)
x² – 4x – 621 = (x + 23)(x – 27)

(vi) x² – 10x – 171
= (x)² – 10x + 25 – 25 – 171
= (x)² – 2 . x . 5 + (5)² – 196
= (x – 5)² – (14)²
= (x – 5 + 14)(x – 5 – 14)
= (x + 9)(x – 19)
x² – 10x – 171 = (x + 9)(x – 19)

(vii) x² – 6x – 891
= x² – 6x + 9 – 9 – 891
= {(x)² – 2 . x . 3 + (3)²} – 900
= (x – 3)² – (30)²
= (x – 3 + 30)(x – 3 – 30)
= (x + 27)(x – 33)
x² – 6x – 891 = (x + 27)(x – 33)

(viii) x² + 4x – 192
= x² + 4x + 4 – 4 – 192
= (x)² + 2 . x . 2 + (2)² – 196
= (x + 2)²-(14)²
= (x + 2 + 14)(x + 2-14)
= (x + 16)(x – 12)
x² + 4x – 192 = (x + 16)(x – 12)