Math Master

∴ x - 2, f(x) = x² - x - a এর একটি উৎপাদক।
∴ f(2) = 2² - 2 - a = 0
বা, 4 - 2 - a = 0
বা, 2 - a = 0
∴ a = 2

x + 1 দ্বারা f(x) = 3x³ + 2x² - 21x - 30 বিভাজ্য
∴ ভাগশেষ f(-1) = 3(-1)³ + 2(-1)² - 21(-1) - 30
= -3 + 2 + 21 - 30
= 23 - 33
= -10

x² + 7x - 60 = x2 + 12x - 5x - 60
= x(x + 12) - 5(x + 12)
= (x + 12)(x - 5)

2x³ + 5x² - 6x - 16
= 2x³ + 4x² + x² + 2x - 8x - 16
= 2x²(x + 2) + x(x + 2) - 8(x + 2)
= (x + 2)(2x² + x - 8)

a² - c² + 2bc - b²
= a² - (b² - 2bc + c²)
= a² - (b - c)²
= (a + b - c)(a - b + c)

x² + x - (a + 1)(a + 2)
= x² + x - (a + 1){(a + 1) + 1}
= x² + x - b(b + 1) [ধরি, b = a + 1]
= x² + x - b² - b
= x² + x - b² - b
= x² - b² + x - b
= (x + b)(x - b) + 1(x - b)
= (x - b)(x + b + 1)
= (x - a - 1)(x + a + 1 + 1)
= (x - a - 1)(x + a + 2)

4x⁴ + 16
= 4(x⁴ + 4)
= 4[(x²)² + 2²]
= 4[(x² + 2)² - 2.x².2]
= 4[(x² + 2)² - (2x)²]
= 4(x² + 2x + 2)(x² - 2x + 2)

x⁶ - 1
= (x³ + 1)(x³ - 1)
= (x + 1)(x² - x + 1)(x - 1)(x² + x + 1)

y² + 2xy - 2x - 1
= y² - 1 + 2xy - 2x
= (y + 1)(y - 1) + 2x(y - 1)
= (y - 1)(2x + y + 1)

f(x) = 3x² - 7x - 16 কে 3x + 2 দ্বারা ভাগ করলে
ভাগশেষ f{-(2/3)} = 3{-(2/3)}² - 7{-(2/3)} - 16
= 4/3 + 14/3 - 16
= (4 + 14 - 48)/3
= -30/3
= -10

এখানে,
f(x) = x³ - 6x² + 11x - 6
f(4) = 64 - 96 + 44 - 6
= 6
∴ x - 4, f(x) এর উৎপাদক নয়।

y² - x(x - 2) - 1
= y² - x² + 2x - 1
= y² - (x² - 2x + 1)
= y² - (x - 1)²
= (y + x - 1)(y - x + 1)
= (x + y - 1)(y - x + 1)

x⁴ + x² + 1
= (x²)² + 2.x².1 + 1 - x²
= (x² + 1)² - x²
= (x² + 1 + x)(x² + 1 - x)
= (x² + x + 1)(x² - x + 1)

3x² - 3x - xy + y
= 3x(x - 1) - y(x - 1)
= (x - 1)(3x - y)

x³ - 729
= x³ - 9³
= (x - 9)(x² + 9x + 81)

4x⁴ - 25x² + 36
= (2x²)² - 2.2x².6 + 6² - x²
= (2x² - 6)² - x²
= (2x² + x - 6)(2x² - x - 6)
= (2x² + 4x - 3x - 6)(2x² - 4x + 3x - 6)
= [2x(x + 2) - 3(x + 2)][2x(x - 2) + 3(x - 2)]
= (x + 2)(2x - 3)(x - 2)(2x + 3)

f(-5) = (-5)² + 7(-5) + a
= 25 - 35 + a
= a - 10
∴ f(-5) = 0 হলে,
a - 10 = 0
∴ a = 10

xy + 2ay + 3ax + 6a²
= y(x + 2a) + 3a(x + 2a)
= (x + 2a)(y + 3a)

x + 1, দ্বারা f(x) নিঃশেষে বিভাজ্য হলে f(-1) = 0
বা, a(-1)² - (a - 1)(-1) - (a + 2) = 0
বা, a + a - 1 - a - 2 = 0
বা, a - 3 = 0
∴ a = 3

x⁶ + 4x³ - 1
= x⁶ + x³ - 1 + 3x³
= (x²)³ + x³ + (-1)³ - 3.x².x(-1)
= (x² + x - 1){(x²)² + x² + (-1)² - x².x - x²(-1) - x(-1)}
= (x² + x - 1)(x⁴ + x² + 1 - x³ + x² + x)
= (x² + x - 1)(x⁴ - x³ + 2x² + x + 1)