Question

A quadratic polynomial is exactly divisible by (x + 1) & (x+2) and leaves the remainder 4 after division by (x +
3) then that polynomial is

(A) x2 + 6x + 4 (B) 2x2 + 6x + 4 (C) 2x2 + 6x - 4 (D) x2 + 6x - 4​

Answer 1

Given:

1) A quadratic equation is exactly divisible by

(x + 1) and (x +2)

2) It leave remainder 4 when divided by (x + 3).

Solution:

Let the required quadratic equation be

ax² + bx + c = 0.

i.e. f (x) = ax² + bx + c

According to first condition.

f (x) = ax² + bx + c

When divided by (x + 1), by using remainder theorem

f (-1) = 0

》a (-1)² + b (-1) + c = 0

》》a - b + c = 0 ...(1)

When divided by (x + 2), by using remainder theorem

f (-2) = 0

》a (-2)² + b (-2) + c = 0

》》4a - 2b + c = 0 ...(2)

According to the second condition.

f (x) = ax² + bx + c

When divided by (x + 3), by using remainder theorem

f (-3) = 4

》a (-3)² + b (-3) + c = 4

》》9a - 3b + c = 4 ...(3)

Subtract eq (2) from eq (3), we get

》》5a - b = 4 ...(4)

Subtract eq (1) from eq (2), we get

》》3a - b = 0 ...(5)

Subtract eq (5) from eq (4), we get

》2a = 4

》a = 2

Substitute a = 2 in eq (5), we get

》3 (2) - b = 0

》b = 6

Substitute a = 2 and b = 6 in eq (1), we get

》2 - 6 + c = 0

》c = - 4

Substituting values of a, b and c in

ax² + bx + c = 0 (general form of quadratic equation)

》2x² + 6x - 4 = 0, is the required quadratic equation.