Let p (x) = x3 - 3x2 - 12x + 19 and q (x) = x2 + x - 6
By division algorithm, when p (x) is divided by q (x), the remainder is a linear expression in x.
So, let r (x) = ax + b is added to p (x) so that p (x) + r (x) is divisible by q (x).
Let,
f (x) = p (x) + r (x)
= x3 - 3x2 - 12x + 19 + ax + b
= x3 – 3x2 + x (a – 12) + b + 19
We have,
q (x) = x2 + x - 6
= (x + 3) (x – 2)
Clearly, q (x) is divisible by (x – 2) and (x + 3) i.e. (x – 2) and (x + 3) are factors of q (x)
We have,
f (x) is divisible by q (x)
(x – 2) and (x + 3) are factors of f (x)
From factor theorem,
If (x – 2) and (x + 3) are factors of f (x) then f (2) = 0 and f (-3) = 0 respectively.
f (2) = 0
(2)3 – 3 (2)2 + 2 (a – 12) + b + 19 = 0
⇒ 8 – 12 + 2a – 24 + b + 19 = 0
⇒ 2a + b – 9 = 0 (i)
Similarly,
f (-3) = 0
(-3)3 – 3 (-3)2 + (-3) (a – 12) + b + 19 = 0
⇒ - 27 – 27 – 3a + 36 + b + 19 = 0
⇒ b – 3a + 1 = 0 (ii)
Subtract (i) from (ii), we get
b – 3a + 1 – (2a + b – 9) = 0 – 0
⇒ b – 3a + 1 – 2a – b + 9 = 0
⇒ - 5a + 10 = 0
⇒ 5a = 10
⇒ a = 2
Put a = 2 in (ii), we get
b – 3 × 2 + 1 = 0
⇒ b – 6 + 1 = 0
⇒ b – 5 = 0
⇒ b = 5
Therefore,
r (x) = ax + b
= 2x + 5
Hence,
x3 – 3x – 12x + 19 is divisible by x2 + x – 6 when 2x + 5 is added to it.