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SEBA Class 8 Mathematics Chapter 14 বীজগণিতীয় ৰাশিৰ উৎপাদক বিশ্লেষণ
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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
= 2p3 – 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
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