# 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?

## 2 Answers

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