Koshe dekhi 13.1 Class 8

(x − y)²− x + y − 2
= (x − y)²+ (−1)(x − y) + (−2)
= (x − y)²+ {(−2) + 1}(x − y) + (−2)×1 ≡ X² + (p + q)X + pq
= (x − y − 2)(x − y + 1) ≡ (X + p) (X + q)
∴ p = − 2, q = 1

বীজগাণিতিক সংখ্যামালা	p ও q এর মান	উৎপাদকে বিশ্লেষণ
x²− 40x − 129	p = − 43 , q = 3	(x − 43)(x + 3)
m² + 19m + 60	p = 15 , q = 4	(m + 15)(m + 4)
x²− x − 6	p = − 3 , q = 2	(x − 3)(x + 2)
(a + b)²− 4(a + b) − 12	p = − 6 , q = 2	(a + b − 6)(a + b + 2)
(x − y)² − x + y − 2	p = − 2 , q = 1	(x − y − 2)(x − y + 1)

2. উৎপাদকে বিশ্লেষণ করি

(i) (a + b)²− 5(a + b) − 6

সমাধানঃ
(a + b)²− 5(a + b) − 6
= (a + b)²+ (−5)(a + b) + (−6)
= (a + b)²+ {(−6) + 1}(a + b) + (−6)×1
= (a + b − 6)(a + b + 1) [যেহেতু, x² + (p + q)x + pq = (x + p) (x + q)]

(ii) (x²− 2x)²+ 5 (x²− 2x) − 36

সমাধানঃ
(x²− 2x)²+ 5(x²− 2x) − 36
= (x²− 2x)²+ 5(x²− 2x) + (−36)
= (x²− 2x)²+ {9 + (−4)}(x²− 2x) + {9×(−4)}
= (x²− 2x + 9)(x²− 2x − 4) [যেহেতু, x² + (p + q)x + pq = (x + p) (x + q)]

(iii) (p² − 3q²)² − 16(p² − 3q²) + 63

সমাধানঃ
(p² − 3q²)² − 16(p² − 3q²) + 63
= (p² − 3q²)² + (−16)(p² − 3q²) + 63
= (p² − 3q²)² + {(−9) + (−7)}(p² − 3q²) + (−9)×(−7)
= (p² − 3q² − 9)(p² − 3q² − 7) [যেহেতু, x² + (p + q)x + pq = (x + p) (x + q)]

আরও দেখুনঃ

(iv) a⁴+ 4a² − 5

সমাধানঃ
a⁴+ 4a² − 5
= (a²)² + 4a² + (−5)
= (a²)²+ {5 + (−1)}a² + 5×(−1)
= (a² + 5)(a² − 1) [যেহেতু, x² + (p + q)x + pq = (x + p) (x + q)]
= (a² + 5)(a² − 1²)
= (a² + 5)(a + 1)(a − 1)

(v) x²y² + 23xy − 420

সমাধানঃ
x²y² + 23xy − 420
= (xy)² + 23xy + (−420)
= (xy)²+ {35 + (−12)}xy + {35×(−12)}
= (xy + 35)(xy − 12) [যেহেতু, x² + (p + q)x + pq = (x + p) (x + q)]

(vi) x⁴ − 7x²+ 12

সমাধানঃ
x⁴− 7x² + 12
= (x²)²+ (−7)x² + 12
= (x²)²+ {(−4) + (−3)}x² + (−4)×(−3)
= (x² − 4)(x² − 3) [যেহেতু, x² + (p + q)x + pq = (x + p) (x + q)]
= (x² − 2²)(x² − 3)
= (x + 2)(x − 2)(x² − 3)

(vii) a²+ ab − 12b²

সমাধানঃ
a² + ab − 12b²
= a² + ab + (−12)b²
= a²+ {4 + (−3)}ab + {4×(−3)}b²
= (a + 4b)(a − 3b) [যেহেতু, x² + (p + q)x + pq = (x + p) (x + q)]

আরও দেখুনঃ

(viii) p² + 31pq + 108q²

সমাধানঃ
p² + 31ab + 108q²
= p² + (27 + 4)pq + (27)×(4)q²
= (p + 27q)(p + 4q) [যেহেতু, x² + (p + q)x + pq = (x + p) (x + q)]

(ix) a⁶ + 3a³b³ − 40b⁶

সমাধানঃ
a⁶ + 3a³b³ − 40b⁶
= (a³)² + 3a³b³ + (−40)(b³)²
= (a³)² + {8 + (−5)}a³b³ + {8×(−5)}(b³)²
= (a³ + 8b³)(a³ − 5b³) [যেহেতু, x² + (p + q)x + pq = (x + p) (x + q)]
= {(a)³ + (2b)³}(a³ − 5b³)
= (a + 2b){a² − a×2b + (2b)²}(a³ − 5b³)
= (a + 2b)(a² − 2ab + 4b²)(a³ − 5b³)

(x) (x + 1)(x + 3)(x − 4)(x − 6) + 24

সমাধানঃ
(x + 1)(x + 3)(x − 4)(x − 6) + 24
= {(x + 1)(x − 4)}{(x + 3)(x − 6)} + 24
= [x² + {1 + (−4)x + 1×(−4)][x²+ {3 + (−6)}x + 3×(−6)] + 24
= (x²− 3x − 4)(x²− 3x − 18) + 24
ধরি x²− 3x = a
= (a − 4)(a − 18) + 24
= {a² + {(−4) + (−18)}a + (−4)×(−18)} + 24
= a² + (−22)a + 72 + 24
= a² + (−22)a + 96
= a² + {(−16) + (−6)}a + (−16)×(−6)
= (a − 16)(a − 6) [যেহেতু, x² + (p + q)x + pq = (x + p) (x + q)]
এখন, a -এর মান অর্থাৎ x² − 3x বসিয়ে পাই-
= (x² − 3x − 16)(x² − 3x − 6)

(xi) (x + 1)(x + 9)(x + 5)²+ 63

সমাধানঃ
(x + 1)(x + 9)(x + 5)²+ 63
= {(x + 1)(x + 9)}{(x + 5)(x + 5)} + 63
= {x² + (9 + 1)x + 9×1}{x²+ (5 + 5)x + 5×5} + 63
= (x² + 10x + 9)(x²+ 10x + 25) + 63
ধরি x² + 10x = a
= (a + 9)(a + 25) + 63
= a² + (25 + 9)a + 25×9 + 63
= a² + 34a + 225 + 63
= a² + 34a + 288
= a² + (18 + 16)a + 18×16
= (a + 18)(a + 16) [যেহেতু, x² + (p + q)x + pq = (x + p) (x + q)]
এখন, a -এর মান অর্থাৎ x² + 10x বসিয়ে পাই-
= (x² + 10x + 18)(x² + 10x + 16)
= (x² + 10x + 18){x² + (8 + 2)x + 8×2}
= (x² + 10x + 18)(x + 8)(x + 2) [যেহেতু, x² + (p + q)x + pq = (x + p) (x + q)]

আরও দেখুনঃ