What must be added to x³ – 3x² – 12x + 19 so that the result is exactly divisible by x² + x – 6.

Let ‘a’ is added to x³ – 3x² – 12x + 19 so that it is exactly divisible by x² + x – 6

Dividing x³ – 3x² – 12x + 19 + a by x² + x – 6

As x² + x – 6 exactly divides x³ – 3x² – 12x + 19 hence remainder should be 0

Equating remainder to 0

⇒ – 2x – 5 + a = 0

⇒ a = 2x + 5

Hence 2x + 5 should be added to x³ – 3x² – 12x + 19 so that it is exactly divisible by x² + x – 6