SEBA Class 8 Mathematics Chapter 14 বীজগণিতীয় ৰাশিৰ উৎপাদক বিশ্লেষণ

SEBA Class 8 Mathematics Chapter 14 বীজগণিতীয় ৰাশিৰ উৎপাদক বিশ্লেষণ Question Answer, SEBA Class 8 Maths Notes in Assamese Medium, SEBA Class 8 Maths Solutions in Assamese to each chapter is provided in the list so that you can easily browse throughout different chapter Assam Board SEBA Class 8 Mathematics Chapter 14 বীজগণিতীয় ৰাশিৰ উৎপাদক বিশ্লেষণ Notes and select needs one.

SEBA Class 8 Mathematics Chapter 14 বীজগণিতীয় ৰাশিৰ উৎপাদক বিশ্লেষণ

Also, you can read the SCERT book online in these sections Solutions by Expert Teachers as per SCERT (CBSE) Book guidelines. SEBA Class 8 Mathematics Chapter 14 বীজগণিতীয় ৰাশিৰ উৎপাদক বিশ্লেষণ Notes. These solutions are part of SCERT All Subject Solutions. Here we have given SEBA Class 8 Mathematics Chapter 14 বীজগণিতীয় ৰাশিৰ উৎপাদক বিশ্লেষণ Solutions for All Subject, You can practice these here.

বীজগণিতীয় ৰাশিৰ উৎপাদক বিশ্লেষণ

Chapter – 14

অনুশীলনী – 14.1

1. তলৰ ৰাশিবোৰৰ উৎপাদক বিশ্লেষণ কৰা।

(i) 3x²y + 5xy

উত্তৰঃ 3x²y + 5xy

= xy × 3x + xy + 5

= xy (3x + 5)

(ii) 10x²y – 5xy²

উত্তৰঃ 10x²y – 5xy²

= 5xy × 2x – 5xy × y

= 5xy (2x – y) 

(iii) 7a²bc – 21ab²c + 14abc

উত্তৰঃ 7a²bc – 21ab²c + 14abc

= 7abc × a – 7abc × 3b + 7abc × 2 

= 7abc (a – 3b + 2)

2. উৎপাদকত প্ৰকাশ কৰাঃ

(i) a² + ab + 6a + 6b

উত্তৰঃ a² + ab + 6a + 6b 

= a(a + b) + 6(a + b)

= (a + b) (a + 6)

(ii) a² + bc + ab + ac

উত্তৰঃ a² + bc + ab + ac

= a² + ab + bc + ac 

= a(a + b) + c(a + b) 

= (a+b) (a+c)

(iii) 1 + x + x² + x³

উত্তৰঃ 1 + x + x² + x³

= 1 + x + x² (1 + x) 

= (1 + x) + x² (1 + x) 

= (1 + x) (1 + x²)

(iv) ab + a + b + 1

উত্তৰঃ ab + a + b + 1

= a(b + 1) + 1(b + 1)

= (b + 1) (a + 1) 

= (a + 1) (b + 1)

(v) 4ax+3ay-4bx-3by

উত্তৰঃ 4ax + 3ay – 4bx – 3by 

= a(4x + 3y) – b(4x + 3y)

= (a – b) (4x + 3y)

3. উৎপাদকত প্ৰকাশ কৰাঃ

(i) x² – 3y

উত্তৰঃ x² – 36 

= x² – 62 [∵ a² – b² = (a + b) (a – b)]

= (x + 6) (x – 6)

(ii) 9x² + 30x + 25

উত্তৰঃ 9x² + 30x + 25 

= (3x)² + 2(3x).5 + 5²

= (3x + 5)²

= (3x + 5) (3x + 5)

(iii) 16a²- 88a + 121

উত্তৰঃ 16a² – 88a + 121

= (4a)² – 2(4a) × 11 + (11)²

= (4a -11)²

= (4a -11) (4a -11)

(iv) 11x²- 44

উত্তৰঃ 11x² – 44 

= 11(x²- 4)

= 11(x² – 2²)

= 11(x + 2) (x – 2)

(v) x⁴ – 81

উত্তৰঃ x⁴ – 81

= (x²)² – 9²

= (x² + 9) (x² – 9)

= (x² + 9) {x² – (3)²} 

= (x² + 9) (x + 3) (x – 3)

(vi) 4 – x² – y² + 2xy

উত্তৰঃ 4 – x² – y² + 2xy 

= 4 – (x² + y² – 2xy)

= 4 – (x² – 2xy + y²)

= 2² – (x – y)²

= {2 + (x-y)} {2 – (x – y)}

= (2 + x – y) (2 – x + y)

(vii) x⁸ – y⁸ 

উত্তৰঃ x⁸ – y⁸

= (x⁴)² – (y⁴)²

= (x⁴ + y⁴) (x⁴ – y⁴)

= (x⁴ + y⁴) {(x²)² – (y²)²}

= (x⁴ + y⁴) (x² + y²) (x² – y²) 

= (x⁴ + y⁴) (x² + y²) (x + y) (x − y)

= (x − y) (x + y) (x² + y²) (x⁴ + y⁴)

(viii) a³ – ab² – a²b + b³

উত্তৰঃ a³ – ab² – a²b + b³

= a(a² – b²) – b(a² – b²)

= (a – b) (a² – b²)

= (a-b) (a – b) (a+b)

4. উৎপাদকত প্ৰকাশ কৰাঃ

(i) 16 + 8x + x²

উত্তৰঃ 16 + 8x + x²

= x² + 8x + 16

= x² + 2x(4) + (4)²

= (x + 4)²

= (x + 4) (x + 4)

(ii) 15 – 2x – x²

উত্তৰঃ 15 – 2x – x² = -[x² + 2x-15] (mn² + px + q আৰ্হিত সজাই লৈ) 

= -[x² + 5x – 3x – 15] 

= -[x(x + 5) – 3(x + 5)] 

= -[(x + 5) (x – 3)] 

= -(x + 5) (3 – x)

= (x + 5) (x – 3)

(iii) x² + 8x – 20

উত্তৰঃ x² + 8x – 20 

= x² + 10x – 2x – 20

= x(x + 10) – 2(x + 10)

= (x + 10) (x – 2)

(iv) x² + 2x – 3

উত্তৰঃ x² + 2x – 3 

= x² + 3x – x – 3

= x(x + 3) – 1(x + 3) 

= (x + 3)(x – 1)

(v) a² – 4a – 12

উত্তৰঃ a² – 4a – 12 

= a² – 6a + 2a – 12 

= a(a – 6) + 2(a – 6) 

= (a + 2) (a – 6)

(vi) x² – 21x + 104 

উত্তৰঃ x² – 21x + 104

= x² – 8x – 13x + 104

= x(x – 8) – 13(x – 8) 

= (x – 8) (x -13)

(vii) 2x² + 18x + 40  

উত্তৰঃ 2x² + 18x +40

= 2[x² + 9x + 20] 

= 2[x² + 5x + 4x + 20]

= 2[x(x + 5) + 4(x + 5)]

= 2(x+5) (x+4) 

= 2(x+4) (x+5)

(viii) l² -13l + 42

উত্তৰঃ l² – 13l² + 42

= l² – 7l – 6l + 42 

= l(l – 7) – 6 (l – 7) 

= (l – 7) (l – 6)

(ix) -a² – a + 20 

উত্তৰঃ -a² – a + 20

= -[a² + a – 20]

= -[a² + 5a – 4a – 20]

= -[a(a + 5) – 4(a + 5)] 

= -[(a + 5) (a – 4)]

= -(a + 5) (a – 4)

= (4 – a) (a + 5)

5. তলৰ ৰাশিবোৰৰ উৎপাদক বিশ্লেষণ কৰাঃ

(i) 3x² + 8x + 4

উত্তৰঃ 3x² + 8x + 4 

= 3x² + 6x + 2x + 4

= 3x(x + 2) + 2(x + 2)

= (x + 2) (3x + 2)

(ii) 2m² + 7m + 3

উত্তৰঃ 2m² + 7m + 3

= 2m² + 6m + m + 3

= 2m (m + 3) + 1(m + 3)

= (m + 3) (2m + 1)

(iii) 2p²+p-28

উত্তৰঃ 2p² + p – 28 

= 2p² + 8p – 7p – 28

= 2p (p + 4) -7(p + 4)

= (p + 4) (2p – 7)

(iv) 9a² + 21a – 8

উত্তৰঃ 9a² + 21a – 8

= a² + 24a – 3a – 8 

= 3a (3a + 8) -1(3a + 8)

= (3a + 8) (3a – 1)

(v) 4y² + 25y – 21

উত্তৰঃ 4y² + 25y – 21

= 4y² + 28y – 3y – 21 

= 4y (y + 7) -3 (y + 7)

= (4y – 3) (y + 7)

(vi) 3m⁶ – 6m⁴n – 45m²n²

উত্তৰঃ 3m⁶ – 6m⁴n – 45m²n²

= 3m²[m⁴ – 2m²n – 15n²]

= 3m²[(m²)² -5m²n + 3m²n – 15n²] 

= 3m² [m² (m² – 5n) + 3n (m² – 5n)]

= 3m² (m² – 5n) (m² + 2n)

(vii) 1 – x – 6x²

উত্তৰঃ 1 – x – 6x²

= -[6x² + x – 1] 

= -[6x² + 3x – 2x – 1]

= -[3x(2x + 1) – 1(2x + 1)] 

= -[(3x – 1) (2x + 1)]

= -(3x – 1) (2x + 1)

= (1 – 3x) (2x + 1)

(viii) 6x² + 7xb – 3b²

উত্তৰঃ 6x² + 7xb – 3b²

= 6a² + 9ab – 2ab – 3b²

= 3a (2a + 3b) – b (2a + 3b) 

= (2a + 3b) (3a – b)

6. খালী ঠাই পূৰোৱা (নিৰীক্ষণ কৰি)

(i) 9x² + 15x + 4 = (3x +………) (………+ 1) 

উত্তৰঃ 

[9x² + 15 + 4 = 9x² + 12x + 3x + 4

= 3x (3x + 4) + 1 ( 3x + 4 )

= (3x + 4) (3x + 1)]

(ii) 12y² – 17y + 6 = (………- 2) (4y -………)

উত্তৰঃ 

[12y² – 17y + 6 = 12y² – 9y – 8y + 6

= 3y(4y – 3) – 2(4y – 3)

= (3y – 2) (4y – 3)]

(iii) 6m² – m – 15 = (3m………) (2m………)

উত্তৰঃ 

[6m² – m – 15 = 6m² – 10m + 9m – 15

= 2m (3m – 5) + 3(3m – 5) 

= (3m – 5) (2m + 3)]

অনুশীলনী 14.2

1. তলৰ হৰণবোৰ কৰাঃ

(i) x⁵ ÷ x²

উত্তৰঃ x⁵ ÷ x² = x⁵/x² [aᵐ ÷ aⁿ = aᵐ⁻ⁿ, বিধি ব্যৱহাৰ কৰি] 

= x⁵⁻²

= x³

(ii) 6p³ ÷ 3p

উত্তৰঃ 6p³ ÷ 3p

= 6p³/3p

= 2 × 3p³/3p

= (2p³)/p

= 2p^{3 – 1} 

= 2p²

(iii) 36m³n² ÷ (-4mn³)

উত্তৰঃ 36m³n² ÷ (-4mn³)

= 36m³n²/-4mn³

= 4 × 9m³n²/-4mn³

= 9m³n² /-mn³

= – 9m³⁻¹ .n²⁻³

= – 9m².(n⁻¹)

= – 9m² × 1/n

= – (- 9m² )/n

(iv) 96p²q²r⁴ ÷ 72pqr

উত্তৰঃ 96p²q²r⁴ ÷ 72pqr

(v) -12a⁸b⁷ ÷ 17a⁴b⁹

উত্তৰঃ -12a⁸b⁷ ÷ 17a⁴b⁹

2. তলৰ বহুপদ ৰাশিবোৰক একপদ ৰাশিৰে হৰণ কৰাঃ

(i) (5y³ – 3y²) ÷ y²

উত্তৰঃ (5y³ – 3y²) ÷ y² (5y – 3)

∴ (5y³ – 3y²) ÷ y² = (5y³ – 3y²)/y²

= y² (5y – 3)/y²

= 5y – 3

(ii) (5a⁸ – 4a⁶ + 3a⁴) ÷ 2a⁴

উত্তৰঃ (5a⁸ – 4a⁶ + 3a⁴) ÷ 2a⁴

5a⁸ – 4a⁶ + 3a⁴ = a⁴ (5a⁴ – 4a² + 3)

∴ (5a⁸ – 4a⁶ + 3a⁴) ÷ 2a⁴

(iii) (5p²q³r⁴ – 10p²q²r² + 15p³q³r⁴) ÷ 5p²q²r²

উত্তৰঃ (5p²q³r⁴ – 10p²q²r² + 15p³q³r⁴) ÷ 5p²q²r²

(iv) (ax³ + bx²- cx) ÷  ax

উত্তৰঃ (ax³ + bx²- cx) ÷  ax

(v) (m³n⁶ – m⁶n³) ÷ m³n³

উত্তৰঃ (m³n⁶ – m⁶n³) ÷ m³n³

3. তলৰ হৰণবোৰ কৰাঃ

(i) (9x – 21) ÷ (3x – 7)

উত্তৰঃ (9x – 21) ÷ (3x – 7)  

= 9x – 21/3x – 7

= 3(3x – 7) )/(3x – 7)

= 3

(ii) 10m(8m + 12) ÷ (4m + 6)

উত্তৰঃ 10m(8m + 12) ÷ (4m + 6)

(iii) 7p²q² (22p-6) ÷ pq(121p – 33)

উত্তৰঃ 7p²q² (22p-6) ÷ pq(121p – 33)

(iv) 1729xyz (3x + 12) (4y – 24) ÷ 19(x + 4)(y – 6)

উত্তৰঃ 1729xyz (3x + 12) (4y – 24) ÷ 19(x + 4)(y – 6)

4. হৰণফল নিৰ্ণয় কৰাঃ

(i) (x² – 25) ÷ (x + 5)

উত্তৰঃ (x² – 25) ÷ (x + 5)

= x² – 25/x + 5

= x² – 5²/x + 5

= (x + 5)(x – 5)/(x + 5)

= (x – 5)

(ii) (4a² + 8a + 4) ÷ (a + 1)²

উত্তৰঃ (4a² + 8a + 4) ÷ (a + 1)²

(iii) (9p² – 18p + 9) ÷ (p -1)

উত্তৰঃ (9p² – 18p + 9) ÷ (p -1)

(iv) 26pqr (p + q) (q + r) (r + p) ÷ 52pq (q + r) (r + p)

উত্তৰঃ 26pqr (p + q) (q + r) (r + p) ÷ 52pq(q + r)(r + p)

= 26pqr (p + q) (q + r) (r + p)/52pq (q + r)(r + p)

= 26pqr (p + q)/52pq

= 26r (p + q)/(2 × 26)

= r (p+q)/2

(v) (x⁴ – 81) ÷ (3 – x) 

উত্তৰঃ (x⁴ – 81) ÷ (3 – x)

= (x⁴ – 81)/(3 – x)

= (x²)² – 9²/3 – x

= (x² + 9) (x² – 9)/3 – x

= (x² + 9) (x² – 3²)/3 – x

= (x² + 9)(x + 3)(x – 3)/(3 – x)

= (-x² + 9)(x + 3)(3 – x)/(3 – x)

= -(x² + 9) (x + 3)

(vi) (x² + 10x + 21) ÷ (x + 3)

উত্তৰঃ (x² + 10x + 21) ÷ (x + 3)

= x² + 10x + 21/(x+3)

= x² + 7x + 3x + 21/x + 3

= x(x + 7) + 3(x + 7)/x + 3

= (x + 7)(x + 3)/x + 3

= x + 7

(vii) (m² + 6m – 27) ÷ (m – 3) 

উত্তৰঃ (m² + 6m – 27) ÷ (m – 3)

(viii) (4y² + 25y – 21) ÷ (y + 7) 

উত্তৰঃ (4y² + 25y – 21) ÷ (y + 7)

(ix) (4u² + 25u + 21) ÷ (u +1)

উত্তৰঃ (4u² + 25u + 21) ÷ (u +1)

(x) 52y³ (50y² – 98) ÷ 26y²(5y+7) 

উত্তৰঃ 52y³ (50y² – 98) ÷ 26y²(5y + 7)

= 52y³ (50y² – 98)/26y²(5y + 7)

= 52y³ × 2 (25y² – 49)/26y²(5y + 7)

5. তলৰ গাণিতিক উক্তিবোৰৰ পৰা ভূলটো বাছি উলিওৱা আৰু ভুল বিলাক শুদ্ধ কৰা। 

(i) 9x²/9x² = 0

উত্তৰঃ 

(ii) 4x² + 1/4x² = 1 + 1 = 2

উত্তৰঃ

(iii) (3x+2)/3x=1/2

উত্তৰঃ 3x + 2/3x = 3x/3x + 2/3x

= 1 + 2/3x

∴ 3x + 2/3x = 1 + 2/3x

(iv) (7x + 5)/5 = 7x

উত্তৰঃ 7x + 5/5 = 7x/5 + 5/5

= 7x/5 + 1

∴ 7x + 5)/5 = 7x/5 + 1

(v) 4x² + 8x + 4/4 = x² + 8x + 4 

উত্তৰঃ 4x² + 8x + 4/4 = 4(x² + 2x + 1)/4

= x² + 2x + 1

∴ 4x² + 8x + 4/4 = x² + 2x + 1