What must be added to x^3 - 3x^2 - 12x + 19 so that the result is exactly divisible by x^2 + x - 6

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.

Let  p(x) =  x3 − 3x2 − 12x + 19 and `q(x) = x^2 + x -6 `be the given polynomial.

When p(x) is divided by q(x), the reminder is a linear polynomial 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)

Then,

`f(x) = x^3 -3x^2 -12x + 19 +ax + b`

` = x^3 -3x^2 + (a-12)x+ (19+b)`

We have,

`q(x) = x^2 + x-6`

       ` = (x +3)(x - 2)`

Clearly, q(x) is divisible by  (x+3)and  (x- 2)i.e.,  (x+3) and (x-2) are the factors of q(x).

Therefore, f (x) is divisible by q(x), if (x + 3) and  (x-2)are factors of f(x), i.e.,

f(-3)and f(2) = 0
Now, f(-3) = 0

\[\Rightarrow\] f(-3) = (-3)3 -3(-3)2 + (a-12)(-3)+19+b = 0

\[\Rightarrow\]  -27 - 27 - 3a + 36 + 19 + b = 0

\[\Rightarrow\] -27 - 27 - 3a + 36 + 19 + b = 0\[\Rightarrow\] -54 - 3a + b + 55 = 0

\[\Rightarrow\]  -3a + b + 1 = 0    ---- (i)

And

`f(2) = (2)^3 -3(2)^2 + (a-12) + 19 +b = 0`

`8-12+2a - 24 +19 +b = 0`

                                    `2a +b = 9            ............. (2)`

Subtracting (i) from (ii), we get,

`(2a +b)-(-3a + b) = 10`

                                     `5a = 10`

                                        `a=2`

Putting this value in equation (ii), we get,

`⇒ 2 xx 2 +b = 9`

                      `b = 5`

Hence, p(x) is divisible by q(x) if  (2x + 5)added to it.