1. ## Factorizing cubics

Hi,

Is there some kind of formula for factorizing cubic s?

By formula i mean. If we have a second degree polynomial $\displaystyle x^2+6x+12$ we know we need two numbers that multiply to give us 12 and add together to give us 6. Is there any similar forumula for cubics?

Im having some difficulties factorizing the expression $\displaystyle x^4-11x^3-36x^2-36$ i know that 2 is a factor so after taking out x-2 i end up with $\displaystyle x^3-9x^2+18$

The closest i got was $\displaystyle (x-6x)(x^2-3x)$ doesn't really cut it though =/

2. Actually, for that quartic, 2 is not a factor.

It has 2 real and 2 complex solutions. The real ones are not nice.

You do mean $\displaystyle x^{4}-11x^{3}-36x^{2}-36$

Make sure you did not make a typo with a sign.

3. Originally Posted by galactus
Actually, for that quartic, 2 is not a factor.

It has 2 real and 2 complex solutions. The real ones are not nice.

You do mean $\displaystyle x^{4}-11x^{3}-36x^{2}-36$

Make sure you did not make a typo with a sign.

Hmm, you're right, i must have missed(or added) a parenthesis somewhere =/

But anyway is there an easy way of solving these equations?
i know the usual trick won't work here since we have an $\displaystyle x^3$ term

Otherwise you could set $\displaystyle x^2=t$

4. Originally Posted by Jones
Hmm, you're right, i must have missed(or added) a parenthesis somewhere =/

But anyway is there an easy way of solving these equations?
i know the usual trick won't work here since we have an $\displaystyle x^3$ term

Otherwise you could set $\displaystyle x^2=t$
If a polynomial with integer coefficients has "nice" factors (that is the polynomial equation that you get by setting the polynomial to zero has rational roots) the rational roots theorem tell us they are the ratio of a factor of the constant term to a factor of the coefficient of the highest power of the variable.

$\displaystyle x^{4}-11x^{3}-36x^{2}-36$

the candidates for rational roots are $\displaystyle \pm 1,\ \pm 2,\ \pm 3,\ \pm 6,\ \pm 9,\,\ \pm 12,\ \pm 18,\ \pm 36$ we try all of these and find that none of them is a root, so this quartic has no nice factors.

RonL

5. So, um.. how would you solve something like that?

6. Originally Posted by Jones
So, um.. how would you solve something like that?
Graphically or numerically. The point is that your question is flawed somehow.

-Dan

7. Originally Posted by Jones
So, um.. how would you solve something like that?
Depends; numerical and graphical solutions are often adequate, but the is an algebraic method which for most practical purposes is useless, but see here.

RonL