Question

What must be subtracted from x³ − 6x² − 15x + 80 so that the result is exactly divisible by x² + x − 12?

Answer 1

By divisible algorithm, when  `p(x) = x^3 - 6x^2 - 15x + 80` is divided by  `x^2 + x -12` the reminder is a linear polynomial

Let r(x) = a(x) + b be subtracted from p(x) so that the result is divisible by q(x).

Let

`f(x) = p(x) -q(x)`

` = x^3 - 6x^2 - 15x + 80 - (ax + b)`

` = x^3 -6x^2 - (a+15)x + 80 -b`

We have,

`q(x) = x^2 + x - 12`

` = x^2 + 4x -3x -12`

` = (x+4)(x-3)`

Clearly, (x+4)and (x-3)are factors of q(x), therefore, f(x) will be divisible by q(x) if (x+4) and (x-3) are factors of f(x), i.e. f (−4) and f (3) are equal to zero.

Therefore,

`f(-4) = (-4)^3 -6(-4)^2 - (a+15)(-4) + 80 -b = 0`

                            `-64 -96 + 4a + 60 + 80 -b = 0`

                                                         ` -20 +4a -b = 0`

                                                                  `+4a -b = 20`

and

`f(3) = (3)^2 - (a+15)(3) + 80 -b = 0`

`27 - 25 - 3a - 45 + 80 -b =0`

                                    ` -3a - b = 8`

                                        `3a +b = 8`

Adding (i) and (ii), we get,

 a=4

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

b=-4

Hence, `x^3 - 6x^2 - 15x + 80` will be divisible by `x^2 + x - 12,` if 4 x − 4 is subtracted from it.