Question

How do you divide `(3x^3-6x^2+13x-4)/(x-3)`?

Answer 1

`(3x^3-6x^2+13x-4)/(x-3)=(3x^2+3x-22)+62/(x-3)`

Explanation:

We want to evaluate `(3x^3-6x^2+13x-4)/(x-3)`

First let `f(x)=3x^3-6x^2+13x-4`

Now, we'll take `f(3)` to see if `(x-3)` is a factor of `f` or not

`f(3)=62`

So `(x-3)` isn't a factor.

But what can say is:

`3x^3-6x^2+13x-4=(x-3)(ax^2+bx+c)+62`

`3x^3-6x^2+13x-66=(x-3)(ax^2+bx+c)`

`therefore-3c=-66 rArrc=22`
and `ax^3=3x^3 rArr a =3`
and `-9x^2+bx^2=-6x^2 rArrb=3`

`therefore 3x^3-6x^2+13x-4=(x-3)(3x^2+3x-22)+62`

Dividing everything by `(x-3)` gives us

`(3x^3-6x^2+13x-4)/(x-3)=(3x^2+3x-22)+62/(x-3)`