# Factoring a polynomial of degree 4

I am attempting to factor a relatively benign looking polynomial of degree 4. I have tried to use synthetic division to factor this. I was hoping to be able to get a remainder of zero at some point. I don't quite know how this works but I suspect that if perhaps I extend things to complex numbers I might have some luck. For this same reason, I am suspecting that all of the roots are may be complex. Can someone tell me how one deals with this?

In general, something like synthetic division along with the Rational Root Theorem might not be the worst idea, assuming there is a real root. Otherwise, it might be a product of two irreducible quadratic polynomials in which case you may want to look at something like factor-by-grouping. In the worst scenario, you have an 'exotic' polynomial and will have to solve for the roots. Luckily, mathematicians in the 1500s managed to find formulas - albeit of the headache inducing variety - to do exactly this, see the Wiki page for quartic functions.

• I indeed ran across that link, and was considering depressing the quartic or something. In fact just to confirm I am pursuing the correct avenues, a specific case of the polynomial in question is $$(2t^4 - 6 t^3 + 3 t^2 - 12 t -3)$$ – user543213 Mar 19 '18 at 2:14
• @Cows You could use Descartes Rule of Signs to observe that you have a pair of complex roots and a pair of real roots. Using Mathematica to solve the equation, I can tell you that none of them are 'nice'. – mathematics2x2life Mar 19 '18 at 2:30
• I tried something in Mathematica, earlier but am not quite sure. I think i have to tell it to factor and extend over C or something. I attempted using open source tools for this like sympy and so forth but without success. Can you tell me the code you used to achieve this? I got something slightly nicer looking in matlab. – user543213 Mar 19 '18 at 2:49
• @Cows It is as simple as Solve[2 t^4 - 6 t^3 + 3 t^2 - 12 t - 3 == 0, t] or perhaps NSolve[2 t^4 - 6 t^3 + 3 t^2 - 12 t - 3 == 0, t]. The same works for WolframAlpha. – mathematics2x2life Mar 19 '18 at 3:08

Using a CAS can greatly reduce the headache.

• I attempted using CAS, but I am getting some very ugly looking things, I am thinking, may be I was hoping there is a human rendered factorization that abstracts away the ugliness of the factorization. I am also not sure if i did it right – user543213 Mar 19 '18 at 2:46
• CAS's are also very useful for checking results, so if you think you have found a simpler-looking solution, use a CAS to see it actually works. – Grant M Mar 19 '18 at 2:57