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earboth · #2 November 10th, 2008, 07:24 AM

Quote:

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:

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

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