Originally Posted by

**RBry** ...

The next one is to find all the solutions (real and complexe) for:

4x^3 - 4x^2 + 3x - 3 = 0

And, one more using long division: (6x^3 + x^2 - 1) / (x-2)

...

You can factor the LHS of the equation:

$\displaystyle 4x^3 - 4x^2 + 3x - 3 = 0~\implies~4x^2(x-1)+3(x-1)=0~\implies~(x-1)(4x^2+3)=0$

I'll leave the rest for you.

Long division: Code:

51
(6x^3 + x^2 - 1) ÷ (x - 2) = 6x^2 + 13x + 26 + ------
-(6x^3 - 12x^2) x - 2
--------------------
13x^2 + 0*x
-(13x^2 - 26x)
----------------
26x - 1
-(26x - 52)
---------------
51