How to find roots of a 4th order polynomial?

Question

Given that $y=14+z$ and $y=z^4$ find the value of $y$.

Substituting $y=z^4$ we get:

$z^4 = 14 + z$ $\Rightarrow$ $z^4 - z - 14 = 0 $

I do not know how to approach solving this polynomial. The solutions I am looking for are real positive values of $y$.

Answer 1

I could tell by eye that $z=2$ is a solution.

(Looking at divisors of $14$ would be a more formal approach to find that.)

Now $(z^4-z-14)/(z-2)=z^3+2z^2+4z+7$,

which is $7$ when $z=0$ and strictly increasing for $z>0$,

so there are no other real positive solutions.

Correction in response to comment:

I showed that there are no other real positive solutions for $z$, but the question asked for real positive solutions for $y$, and there is another (irrational) one: $z^3+2z^2+4z+7 $ is negative $(1)$ when $z=-2$ and positive $(4)$ when $z=-1$, so there is a solution $z_0$ between $-2$ and $-1$, and $y_0=14+z_0$ is positive. Since $z^3+2z^2+4z+7$ is strictly increasing, there are no other real zeroes of $z^4-z-14$.

Answer 2

There is a formula for quartic equations, but it is very complicated and pretty useless. Have you tried substituting $z$ for some divisor of $14$?

It has no more rational roots. It has no more positive roots, either.