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

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Answer 1 · 2017-06-03T19:55:20

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

We want to evaluate $\frac{3 x^{3} - 6 x^{2} + 13 x - 4}{x - 3}$ $(3x^3-6x^2+13x-4)/(x-3)$

First let $f (x) = 3 x^{3} - 6 x^{2} + 13 x - 4$ $f(x)=3x^3-6x^2+13x-4$

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

$f (3) = 62$ $f(3)=62$

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

But what can say is:

$3 x^{3} - 6 x^{2} + 13 x - 4 = (x - 3) (a x^{2} + b x + c) + 62$ $3x^3-6x^2+13x-4=(x-3)(ax^2+bx+c)+62$

$3 x^{3} - 6 x^{2} + 13 x - 66 = (x - 3) (a x^{2} + b x + c)$ $3x^3-6x^2+13x-66=(x-3)(ax^2+bx+c)$

$therefore-3c=-66 rArrc=22$
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) 3x3−6x2+13x−4x−3(3x^3-6x^2+13x-4)/(x-3) ?