Question

What must be added to x³ − 3x² − 12x + 19 so that the result is exactly divisibly by x² + x - 6 ?

Answer 1

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.