Question

If x2
– 3x + 2 divides x3
– 6x2+ ax + b exactly, then find the value of ‘a’ and ‘b’

Answer 1

2. Factorise: (a – b)3 + (b – c)3 + (c – a)3

2. Factorise: (a – b)3 + (b – c)3 + (c – a)3 Hint: Put a – b = x, b – c = y and c – a = z so that

2. Factorise: (a – b)3 + (b – c)3 + (c – a)3 Hint: Put a – b = x, b – c = y and c – a = z so that x + y + z = 0 ⇒ x3 + y3 + z3 = 3xyz

2. Factorise: (a – b)3 + (b – c)3 + (c – a)3 Hint: Put a – b = x, b – c = y and c – a = z so that x + y + z = 0 ⇒ x3 + y3 + z3 = 3xyz ⇒ (a – b)3 + (b – c)3 + (c – a)3 = 3(a – b) (b – c) (c – a)

2. Factorise: (a – b)3 + (b – c)3 + (c – a)3 Hint: Put a – b = x, b – c = y and c – a = z so that x + y + z = 0 ⇒ x3 + y3 + z3 = 3xyz ⇒ (a – b)3 + (b – c)3 + (c – a)3 = 3(a – b) (b – c) (c – a)3. Factorise: 14x6 – 45x3y3 – 14y6

2. Factorise: (a – b)3 + (b – c)3 + (c – a)3 Hint: Put a – b = x, b – c = y and c – a = z so that x + y + z = 0 ⇒ x3 + y3 + z3 = 3xyz ⇒ (a – b)3 + (b – c)3 + (c – a)3 = 3(a – b) (b – c) (c – a)3. Factorise: 14x6 – 45x3y3 – 14y6 Hint: Let us put x3 = a and y3 = b so t

hat

hat 14x6 – 45x3y3 – 14y6 = 14a2 – 45ab – 14b2

hat 14x6 – 45x3y3 – 14y6 = 14a2 – 45ab – 14b2 = 14a2 – 49ab + 4ab – 14b2 = (2a – 7b) (7a + 2b)

hat 14x6 – 45x3y3 – 14y6 = 14a2 – 45ab – 14b2 = 14a2 – 49ab + 4ab – 14b2 = (2a – 7b) (7a + 2b) ⇒ 14x6 – 45x3y3 – 14y6 = (2x3 – 7y3) (7x3 + 2y3)

hat 14x6 – 45x3y3 – 14y6 = 14a2 – 45ab – 14b2 = 14a2 – 49ab + 4ab – 14b2 = (2a – 7b) (7a + 2b) ⇒ 14x6 – 45x3y3 – 14y6 = (2x3 – 7y3) (7x3 + 2y3)4. Find the product: (x – 3y) (x + 3y) (x2 + 9y2)

hat 14x6 – 45x3y3 – 14y6 = 14a2 – 45ab – 14b2 = 14a2 – 49ab + 4ab – 14b2 = (2a – 7b) (7a + 2b) ⇒ 14x6 – 45x3y3 – 14y6 = (2x3 – 7y3) (7x3 + 2y3)4. Find the product: (x – 3y) (x + 3y) (x2 + 9y2) Hint: (x – 3y) (x + 3y) (x2 + 9y2) and (x2 – 9y2) (x2 + 9y2) = (x4 – 81y4)

hat 14x6 – 45x3y3 – 14y6 = 14a2 – 45ab – 14b2 = 14a2 – 49ab + 4ab – 14b2 = (2a – 7b) (7a + 2b) ⇒ 14x6 – 45x3y3 – 14y6 = (2x3 – 7y3) (7x3 + 2y3)4. Find the product: (x – 3y) (x + 3y) (x2 + 9y2) Hint: (x – 3y) (x + 3y) (x2 + 9y2) and (x2 – 9y2) (x2 + 9y2) = (x4 – 81y4)5. If x2 – 3x + 2 divides x3 – 6x2 + ax + b exactly, then find the value of ‘a’ and ‘b’

hat 14x6 – 45x3y3 – 14y6 = 14a2 – 45ab – 14b2 = 14a2 – 49ab + 4ab – 14b2 = (2a – 7b) (7a + 2b) ⇒ 14x6 – 45x3y3 – 14y6 = (2x3 – 7y3) (7x3 + 2y3)4. Find the product: (x – 3y) (x + 3y) (x2 + 9y2) Hint: (x – 3y) (x + 3y) (x2 + 9y2) and (x2 – 9y2) (x2 + 9y2) = (x4 – 81y4)5. If x2 – 3x + 2 divides x3 – 6x2 + ax + b exactly, then find the value of ‘a’ and ‘b’ Hint: x2 – 3x + 2 = (x – 1) (x – 2) ⇒ x3 – 6x2 + ax + b is exactly divisible by

hat 14x6 – 45x3y3 – 14y6 = 14a2 – 45ab – 14b2 = 14a2 – 49ab + 4ab – 14b2 = (2a – 7b) (7a + 2b) ⇒ 14x6 – 45x3y3 – 14y6 = (2x3 – 7y3) (7x3 + 2y3)4. Find the product: (x – 3y) (x + 3y) (x2 + 9y2) Hint: (x – 3y) (x + 3y) (x2 + 9y2) and (x2 – 9y2) (x2 + 9y2) = (x4 – 81y4)5. If x2 – 3x + 2 divides x3 – 6x2 + ax + b exactly, then find the value of ‘a’ and ‘b’ Hint: x2 – 3x + 2 = (x – 1) (x – 2) ⇒ x3 – 6x2 + ax + b is exactly divisible by (x – 1) and (x – 2) also i.e. f(1) = 0 and f(2) = 0

hat 14x6 – 45x3y3 – 14y6 = 14a2 – 45ab – 14b2 = 14a2 – 49ab + 4ab – 14b2 = (2a – 7b) (7a + 2b) ⇒ 14x6 – 45x3y3 – 14y6 = (2x3 – 7y3) (7x3 + 2y3)4. Find the product: (x – 3y) (x + 3y) (x2 + 9y2) Hint: (x – 3y) (x + 3y) (x2 + 9y2) and (x2 – 9y2) (x2 + 9y2) = (x4 – 81y4)5. If x2 – 3x + 2 divides x3 – 6x2 + ax + b exactly, then find the value of ‘a’ and ‘b’ Hint: x2 – 3x + 2 = (x – 1) (x – 2) ⇒ x3 – 6x2 + ax + b is exactly divisible by (x – 1) and (x – 2) also i.e. f(1) = 0 and f(2) = 0 Now, f(1) = 0 ⇒ a + b – 5 = 0

hat 14x6 – 45x3y3 – 14y6 = 14a2 – 45ab – 14b2 = 14a2 – 49ab + 4ab – 14b2 = (2a – 7b) (7a + 2b) ⇒ 14x6 – 45x3y3 – 14y6 = (2x3 – 7y3) (7x3 + 2y3)4. Find the product: (x – 3y) (x + 3y) (x2 + 9y2) Hint: (x – 3y) (x + 3y) (x2 + 9y2) and (x2 – 9y2) (x2 + 9y2) = (x4 – 81y4)5. If x2 – 3x + 2 divides x3 – 6x2 + ax + b exactly, then find the value of ‘a’ and ‘b’ Hint: x2 – 3x + 2 = (x – 1) (x – 2) ⇒ x3 – 6x2 + ax + b is exactly divisible by (x – 1) and (x – 2) also i.e. f(1) = 0 and f(2) = 0 Now, f(1) = 0 ⇒ a + b – 5 = 0 and f(2) = 0 ⇒ 2a + b – 16 = 0

hat 14x6 – 45x3y3 – 14y6 = 14a2 – 45ab – 14b2 = 14a2 – 49ab + 4ab – 14b2 = (2a – 7b) (7a + 2b) ⇒ 14x6 – 45x3y3 – 14y6 = (2x3 – 7y3) (7x3 + 2y3)4. Find the product: (x – 3y) (x + 3y) (x2 + 9y2) Hint: (x – 3y) (x + 3y) (x2 + 9y2) and (x2 – 9y2) (x2 + 9y2) = (x4 – 81y4)5. If x2 – 3x + 2 divides x3 – 6x2 + ax + b exactly, then find the value of ‘a’ and ‘b’ Hint: x2 – 3x + 2 = (x – 1) (x – 2) ⇒ x3 – 6x2 + ax + b is exactly divisible by (x – 1) and (x – 2) also i.e. f(1) = 0 and f(2) = 0 Now, f(1) = 0 ⇒ a + b – 5 = 0 and f(2) = 0 ⇒ 2a + b – 16 = 0 solving (i) and (ii) a = 11 and b = – 6