Polynomials bx2 + x + 5 and bx3 – 2x + 5 are divided by polynomial x

1.	Polynomials bx2 + x + 5 and bx3 – 2x + 5 are divided by polynomial x – 3 and the remainders are m and n respectively. If m – n = 0, then find the value of b.
Answer» When polynomial bx² + x + 5 is divided by (x – 3), the remainder is m. ∴ By remainder theorem, Remainder = p(3) = m p(x) = bx² + x + 5 ∴ p(3) = b(3)² + 3 + 5 ∴ m = b(9) + 8 m = 9b + 8 …(i) When polynomial bx³ – 2x + 5 is divided by x – 3 the remainder is n ∴ remainder = p(3) = n p(x) = bx³ – 2x + 5 ∴ P(3) = b(3)³ – 2(3) + 5 ∴ n = b(27) – 6 + 5 ∴ n = 27b – 1 …(ii) Now, m – n = 0 …[Given] ∴ m = n ∴ 9b + 8 = 27b – 1 …[From (i) and (ii)] ∴ 8 + 1 = 27b – 9b ∴ 9 = 18b ∴ b = 1/2

Polynomials bx2 + x + 5 and bx3 – 2x + 5 are divided by polynomial x – 3 and the remainders are m and n respectively. If m – n = 0, then find the value of b.

Answer»

When polynomial bx² + x + 5 is divided by (x – 3), the remainder is m.

∴ By remainder theorem,

Remainder = p(3) = m

p(x) = bx² + x + 5

∴ p(3) = b(3)² + 3 + 5

∴ m = b(9) + 8

m = 9b + 8 …(i)

When polynomial bx³ – 2x + 5 is divided by x – 3 the remainder is n

∴ remainder = p(3) = n

p(x) = bx³ – 2x + 5

∴ P(3) = b(3)³ – 2(3) + 5

∴ n = b(27) – 6 + 5

∴ n = 27b – 1 …(ii)

Now, m – n = 0 …[Given]

∴ m = n

∴ 9b + 8 = 27b – 1 …[From (i) and (ii)]

∴ 8 + 1 = 27b – 9b

∴ 9 = 18b

∴ b = 1/2