In summary, after some factoring, the equation 3x^8-20x^6-12x^4+12x^2+1 can be written as (x^2+1)(3x^6-2x^4+11x^2+1). The second factor is a cubic in x^2 which can be solved using the cubic formula, but the resulting solutions are messy. Factoring this equation by hand is not practical and the use of Eisenstein's criterion is limited. The cubic formula is a lengthy formula for solving cubics and is not worth memorizing due to the complexity of the solutions. This equation arises when finding the second derivative of the curve y=(2-x^2)/(
To give you an idea of what the solutions looks like, I used mathematica to solve it and the six solutions from the cubic formula all look like this
[tex]
x = -\sqrt{\frac{23}{9}-\frac{43 5^{2/3}}{9 \left(2 \left(863+27 i \sqrt{69}\right)\right)^{1/3}}+\frac{43 i 5^{2/3}}{3 \sqrt{3} \left(2 \left(863+27 i \sqrt{69}\right)\right)^{1/3}}-\frac{\left(5 \left(863+27 i \sqrt{69}\right)\right)^{1/3}}{9 2^{2/3}}-\frac{i \left(5 \left(863+27 i \sqrt{69}\right)\right)^{1/3}}{3 2^{2/3} \sqrt{3}}}
[/tex]
how did you factor that what is this cubic formula
You can check to see if it might be factorable over Z with eisenstien's criterion, and I guessed one of the factors (which wasn't difficult because the constant is 1) then used polynomial division. If you had to solve something like that factoring wouldn't really be that practical (unless the coefficients were symmetric or something) but there was an easy factor in this. In fact Eisenstein's can't even be used to check if the second factor can be factored, so this isn't something you would do by hand.
The cubic formula is like the quadratic formula but for cubics. It's very long and not worth memorizing because the answers almost always end up messy.