Question

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

Answer 1

`x^3-3x^2-12x+19` is divisible by
`x^2+x-6` when
`2x+5` is added to it.

Explanation:

Let
`p(x)=x^3-3x^2-12+19` &
`q(x)=x^2+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)`

⇒
`f(x)=x^3-3x^2-12x+19+ax+b`

⇒
`f(x)=x^3-3x^2+x(a-12)+b+19`

We have,

`q(x)=x^2+x-6=(x+3)(x-3)`

Clearly,
`q(x)` is divisible by
`(x-2)` and
`(x+3)`

{
`(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 factors 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`

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`

Subtract (1) from (2)

`b-3a+1-(2a+b-9)=0-0`

⇒
`b-3a+1-2a-6+9=0`

⇒
`-5a+10=0` ⇒
`5a=10` ⇒
`a=2`

Put
`a=2` in equation 2

⇒
`b-3` ×
`2+1=0` ⇒
`b-6+1=0` ⇒
`b-5=0` ⇒
`b=5`

∴
`r(x)=ax+b` ⇒
`r(x)=2x+5`

Hence,
`x^3-3x^2-12x+19` is divisible by
`x^2+x-6` when
`2x+5` is added to it.

Hope this helps...