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

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Answer 1 · 2017-06-03T19:55:20

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

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

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

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