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

1 Answer
Jun 3, 2017

#(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)#