WBBSE 9th Class Math Solutions Chapter 8 উৎপাদকে বিশ্লেষণ
WBBSE 9th Class Math Solutions Chapter 8 উৎপাদকে বিশ্লেষণ
West Bengal Board 9th Class Math Solutions Chapter 8 উৎপাদকে বিশ্লেষণ
West Bengal Board 9th Math Solutions
কযে দেখি 8.1
1. x3 – 3x + 2
মনে করি, f(x) = x3 − 3x + 2
f(I) 13 – 3y1 + 2 = 0
অর্থাৎ x = 1 বসিয়ে f (x)-এর মান শূন্য পাওয়া গেল।
∴ উৎপাদক উপপাদ্যের সাহায্যে বলা যায় যে, (x −1),
f (x)-এর একটি দ্বিপদ উৎপাদক।
∴ x3 – 3x + 2
= x3 – x2 + x2 – 2x + 2
= x2(x−1) + x (x–1) – 2 (x – 2 )
= (x−1) (x2 + 2x – x – 2 )
= (x–1) {x (x + 2) –1 (x+2)}
= (x–1) (x–1) (x+2)
=(x−1)2 (x+2).
∴ x3 – 3x + 2 = (x – 1)2 (x+2)
2. x3 +2x + 3
= x3 + x2 – x2 − x + 3x + 3
= x2(x + 1) – x (x + 1) +3 (x + 1)
= (x+1) (x2 – x + 3 )
∴ x3 + 2x + 3 = (x + 1) (x2 – x + 3 )
(বিকল্প পদ্ধতি)
উৎপাদক উপপাদ্যের প্রয়োগ ছাড়া উৎপাদক বিশ্লেষণ।
প্রদত্ত রাশিমালা : x3 + 2x + 3
= x3 + 1 + 2x + 2
= (x + 1) (x2 – x + 1 ) +2 (x + 1)
= (x + 1) (x2 – x + 1 + 2 )
= ( x + 1) (x2 – x + 3)
∴ x3 + 2x + 3 = (x + 1 (x2 – x + 3)
[এক্ষেত্রে a3 + b3 -এর সূত্র ব্যবহার করা হল।]
3. a3 – 12a – 16
= a3 + 2a2 – 2a2 – 4a – 8a – 16
= a2 (a + 2) – 2a (a + 2) – 3 (a + 2)
[a = – 2 বসালে a3 – 12a – 16 = 0 হয়।]
= (a + 2) (a2 – 2a – 8 ) –
= (a + 2) (a2 – 4a + 2a – 8)
= (a + 2) {a (a-4) + 2 (a –4)}
= (a + 2) (a + 2) (a – 4 )
= (a + 2)2 (a – 4 )
∴ a3 – 12a – 16 = (a + 2)2 (a – 4 )
বিকল্প পদ্ধতি : a3 – 12a – 16
= a3 + 8 – 12a – 24
= (a + 2) (a2 – 2a + 4) – 12 (a + 2)
= (a + 2) (a2 – 2a + 4 – 2)
= (a + 2)(a2 – 2a – 8 )
= (a + 2) (a – 4) (a + 2)
= (a + 2)2 (a – 4)
4. x3 – 6x + 4
= x3 – 2x2 + 2x2 – 4x – 2x + 4
= x2 (x − 2) + 2x (x – 2) – 2 (x – 2) [x=2, হলে x3 – 6x + 4 = 0 হয় উৎপাদক উপপাদ্য অনুসারে]
= (x – 2) (x2 + 2x – 2)
বিকল্প পদ্ধতি : x3 – 6x + 4
= x3 – 8 – 6x + 12
= (x-2) (x² + 2x + 4) – 6 (x-2)
= (x − 2) (x² + 2x + 4- 6)
= (4-2) (x²+ 2x – 2)
∴ x3 – 6x + 4 = (x – 2) (x3 + 2x – 2)
5. x3 – 19x – 30
= x3 + 2x2 – 2x2 – 4x – 15x – 30
= x2 – (x + 2) -2x (x + 2) – 15 (x + 2)
[x – 2, হলে x3 – 19x − 30 = 0 হয়]
=(x + 2) (x2 –2x – 15 )
= (x + 2) (x2 – 5x + 3x – 15)
=(x + 2 {x (x –5) +3 (x – 5)}
= (x + 2) (x – 5) (x + 5 )
x3 – 19x – 30 = (x + 3) (x + 2) (x – 5)
বিকল্প পদ্ধতি : x3 −19x – 30
= x3 + 8 – 19x – 38
= (x + 2) (x2 – 2x + 4) – 19 (x + 2)
=(x + 2) (x2–2x – 15) = (x + 2) (x + 3) (x – 5)
6. 4a³ – 9a² + 3a + 2
= 4a3 – 4a2 – 5a2 + 5a – 2a + 2
= 4a (a – 1) – 5a (a – 1) – 2 (a – 1)
[Θ a = 1 বসালে প্রদত্ত রাশির মান শূন্য হয়।]
= (a – 1) (4a2 – 5a – 2)
∴ 4a3 – 9a2 + 3a + 2 = (a – 1) (4a2 – 5a – 2)
7. x3 – 9x2 + 23x – 15
= x3 – x2 –8x2 +8x + 15x -15
= x2 (x – 1) – 8x (x – 1) + 15 (x – 1)
[x = 1 বসালে প্রয়োগরাশির মান শূন্য হয়, উৎপাদক উপাদান অনুসারে, (x−1) একটি উৎপাদক।]
= (x-1) (x²-8x +15)
= (x−1) (x2 – 5x – 3x+15)
=(x-1) { x (x-5) – 3 (x-5)}
= (x – 1) (x – 5) (x − 3 )
∴ x³-9x² +23x – 15 = (x – 1) (x – 5) (x − 3)
8. 5a³ + 11a² + 4a -2
= 5a3 + 5a2+ 6a2 + 6a – 2a – 2
= 5a2 (a+1)+ 6a (a + 1) – 2 (a+1)
= (a + 1) (5a2 + 6a – 2)
5a3 + 11a2 + 4a – 2 = (a+1) (5a2+6a-2)
9. 2x3 – x2 – 9x + 5
= 2x3 + x2 − 2x2 – x + 10x + 5
= x2 (2x + 1) – x (2 x + 1) + 5 (2x + 1)
Θ [x = – 1/2 বসালে প্রদত্ত রাশির মান শূন্য হয়, গুণনীয়ক বা উৎপাদক উপপাদ্য অনুসারে (2x + 1), প্রদত্ত রাশিমালার একটি উৎপাদক]
= (2x + 1) (x2 − x + 5 )
2x3 – x2 + 9x + 5 = (2x + 1) (x2– x + 5 )
10. 2y³ – 5y² – 19y + 42
= 2y3 – 4y2 – y2 + 2y – 21y + 42
= 2y2 (y – 2) – y(y – 2) – 21 (y −2)
= (y – 2) (2y2 –y – 21 )
= (y – 2) (2y2–7y + 6y-21)
= (y-2) {y (2y-7)+3 (2y-7)}
= (y – 2) (2y –7) (y + 3)
∴ 2y³-5y²-19y + 42 = (y-2) (2y-7)(y + 3)
কষে দেখি 8.2
কষে দেখি 8.3
1. t9-512
= (t³)³-83
= (t³-8) (t6+8t³+64)
= (t−2) (t²+2t+4) (t6+8t³+64)
t9-512=(t-2) (t²+2t+4)(t6+8t³+64)
2. 729p6-q6
= (9p²)³-(q²)³
= (9p²-q²)(81p4+9p²q² +q4)
=(3p-q) (3p+q) {9p²)² + 2 × 9p² × q² + (q²)² – 9p²q²}
=(3p-q) (3p+q){(9p²+ q²)² -(3pq)²}
= (3p – q) (3p + q)
(9p²-3pq+q²) (9p²+ 3pq+q²)
3. 8 (p3)³ + 343
= {2(p-3)}³+7³
= [2 (p-3)+7] [{2(p-3)}² -2 (p-3) × 7+77]
= (2p+1) (4p²-24p + 36 – 14p+42 +49)
= (2p+1) (4p²-38p + 127)
8 (p-3)³ + 343 = (2p+ 1)(4p² – 38p + 127)
কষে দেখি 8.4
∴ 8a3 – 27b3 – 1 – 18ab = (2a – 3b – 1) (4b2 + 2b – 3b + 1)
কষে দেখি 8.5
1. (i) (a+b)²-5a-5b+6
= (a+b)²-5(a+b)+6
= x2-5x+6
= x²-3x-2x+6
= x (x-3)-2(x-3)
=(x-3) (x-2)
= (a+b-3) (a+b-2)
∴ (a+b)2 -5a-5b+6 = (a+b-3) (a+b-2)
(ii) (x+1) (x+2) (3x-1) (3x-4) + 12
= (x+1) (3x-1) (x+2) (3x-4)+12
= (3x+2x-1) (3x²+2x-8)+12
মনে করি 3x2+2x=a
= (a-1) (a-8) + 12.
= a²-9a+9+12
= a²-9a+8+12
= a²-9a+20
= a²-5a-4a+20
= a (a-5) -4 (a-5)
= (a-5) (a-4)= (3x²+2x-5) (3x²+2x-4)
= (3x²+5x-3x-5) (3x²+2x-4)
= (3x+5) (x-1) (3x²+2x-4) [Ans.]
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