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
Make sure you did not make a typo with a sign.
Hi,
Is there some kind of formula for factorizing cubic s?
By formula i mean. If we have a second degree polynomial 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 i know that 2 is a factor so after taking out x-2 i end up with
The closest i got was doesn't really cut it though =/
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.
That means for your cubic:
the candidates for rational roots are we try all of these and find that none of them is a root, so this quartic has no nice factors.
RonL