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

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.

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

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

For this I need to find the possible rational roots. The answers are {-2, -2/3, ±i} (There in the back of my book)

I get 2/3 using Synthetic Substitution, (I substituted the possible rational roots and I almost positive 2/3 was the only one to work) The Possible Rational Roots I got were: ±4/3, ±2/3, ±1/3, ±4, ±1

When I synthetically substituted 2/3, I got: 3x^3+6x^2+3x+6

Can some tell how to get, -2 and ±i from this, Thanks

Also, I need to do this without graphing or using the Quartic or Cubic formulas.

Edit: I figured out that the problem was that I forgot to put ±2 as a possible rational root.
ryan82593

Posts: 2
Joined: Sat May 09, 2009 2:10 am

ryan82593 wrote:When I synthetically substituted 2/3, I got: 3x^3+6x^2+3x+6

Can some tell how to get, -2 and ±i from this

As you noticed after initially posting, the Rational Roots Test, when applied to the above result, provides a list of possible roots which includes $\pm 2$.

Once you remove one more linear factor from the polynomial, you'll be down to a quadratic. You can then apply the Quadratic Formula, which will provide you with the complex-valued solutions.

stapel_eliz

Posts: 1718
Joined: Mon Dec 08, 2008 4:22 pm