# Thread: [SOLVED] Polynomial x and i

1. ## [SOLVED] Polynomial x and i

Hi
I tried to find all values of x, real and complex in 3x^3-11x^2+25x-25=0
but I don't get the answer. I used quadratic X(3x^2-11x+25)-25=0 but that didn’t work.

Do I need to use rational roots test and long division to solve this?

2. Originally Posted by Neverquit
Hi
I tried to find all values of x, real and complex in 3x^3-11x^2+25x-25
but I don't get the answer. I used quadratic X(3x^2-11x+25)-25=0 but that didn’t work.

Do I need to use rational roots test and long division to solve this?

Or you'll have to use the method by Tartaglia- del Ferro, which is a very nasty thing, but sometimes there's no other way: Cubic function - Wikipedia, the free encyclopedia

Tonio

3. Your question, as written, "I tried to find all values of x, real and complex in $3x^3-11x^2+25x-25$" doesn't make any sense. x can be any complex number in that formula. Tonio assumed that you meant to solve $3x^3- 11x^2+ 25x-25= 0$. Is that correct?

4. ## correction

Equation is suppose to be: 3x^3-11x^2+25x-25=0

5. ## Cardanos method for polynomials

Cardanos method (below) looks like the easiest way.
I just have to divide 3x^3-11x^2+25x-25 by the leading coefficient 3 to get x^3-(11/3)x^2+(25/3)x-(25/3)=0.
That gives a=(-11/3),b=(25/3),c=(-25/3).Then solve for p,q,u.

Should this work??

For
the solutions for x are given by

where

6. ## Cardanos method for polynomials

Cardanos method gave me the following complex root solutions:
x=2.8444i+1.222

and

x=-1.7175i+1.222

These solutions look close to the answer 1+2i and 1-2i.... just need to find the real root of 5/3.
To find the real root of 5/3 should I use Cardanos method or real roots test?

Actually, it would have been far simpler to do what you originally suggested: "use rational roots test and long division".

The rational roots test shows that only 5, -5, 5/3, and -5/3 could be rational roots. Trying those, we find that 5/3 is, in fact, a root. That means that (3x- 5) is a factor. Dividing $3x^3- 11x^2+ 25x- 25$ by $3x- 5$ gives a quotient of $x^2- 2x+ 5$ with no remainder.

$3x^3- 11x^2+ 25x- 25= (3x- 5)(x^2- 2x+ 5)$

$x^2- 2x+ 5= x^2- 2x+ 1+ 4= (x- 1)^2+ 4= 0$ so
$(x- 1)^2= -4$ and $x- 1= \pm 2i$, $x= 1\pm 2i$.

The roots of the original equation are 5/3, 1+ 2i, and 1- 2i.

(Though, I am, frankly, impressed that you got such accurate roots using Cardano's method.)