Finding the roots of $x^4+x^3+2x^2+x-109=0$

Question

The given equation is $$x^4+x^3+2x^2+x-109=0$$

I was thinking of using substitution method say $x^4=t^2$, but I wasn't sure what to do for the other terms. Using general quadratic theory we can find sum and product of roots, but I need to find the roots here.

Any suggestions?

Answer 1

The substitution you suggested won’t help, I don’t think. 109 is a prime number, so the usual factorization techniques will not work. Two of the roots are real, and the other two are complex. You can either use one of the formulas for quartic roots, or you can use numerical methods.

If you’re feeling lazy, just type your equation into Wolfram Alpha.

Answer 2

Use resultant.

$Res_x(x^4 + x^3 + 2 x^2 + x - 109, A + x^2 + y x)=\\ -(-11881 - 437 A + 216 A^2 + 3 A^3 - A^4 + (-109 + 329 A - A^2 + A^3) y + (218 - 437 A - 2 A^2) y^2 + (-109 + A) y^3 + 109 y^4)\\$

$Res_y(-11881 - 437 A + 216 A^2 + 3 A^3 - A^4 + (-109 + 329 A - A^2 + A^3) y + (218 - 437 A - 2 A^2) y^2 + (-109 + A) y^3 + 109 y^4,-z + (-109 + A) + 4\cdot 109 y)=\\ 109(-3943817860179 - 126648656084 A + 71401744702 A^2 + 1078224076 A^3 - 332291299 A^4 + (-10360232 + 83547192 A + 872 A^2 + 767368 A^3) z + (308906 - 760820 A - 3494 A^2) z^2 + z^4)\\$

Let linear term of last polynomial equal zero, then

1. get $A$ from cubic equation $-10360232 + 83547192 A + 872 A^2 + 767368 A^3=0$

2. get $z$ from biquadratic $-3943817860179 - 126648656084 A + 71401744702 A^2 + 1078224076 A^3 - 332291299 A^4 + (308906 - 760820 A - 3494 A^2) z^2 + z^4=0$

3. get $y$ from $-z + (-109 + A) + 4\cdot 109 y=0$

4. get $x$ from $A + x^2 + y x=0$