Solving a quintic function for zero I got this  question on my homework and I cannot for the life of me figure out how to solve for $0$.
$$x^5+2x-10=0$$
I have tried this every which way and this is my last resort. Thanks in advanced.
 A: Your polynomial has a unique real solution which you can estimate numerically to be around $1.478$. PARI computes the Galois group to be $S_5$, which means that you can't express this solution with radicals (roots).
A: Differentiate and you will see that the function $f(x) = x^5+2x-10$ is strictly increasing.
It is easy to see that $\lim_{x \to -\infty} f(x) = -\infty$ and
$\lim_{x \to \infty} f(x) = \infty$, so there is exactly one root.
Since $f(0) <0$ and $f(2) >0$ we see that the root lies in $(0,2)$.
You can use bisection to find a numerical approximation to the root.
Using 10 iterations of the bisection method starting with the interval $[0,2]$
gives the bracket $(1.4765625, 1.478515625)$.
A: If $$f(x)=x^5+2x-10$$ $$f'(x)=5x^4+2 >0$$ so there is only one real root since the function varies from $-\infty$ to $+\infty$ because of $x^5$.
By inspection, you can notice that $f(1)=-7$ and $f(2)=+26$ so the solution is between $1$ and $2$.
To find the solution, Newton method is quite simple : starting from a "reasonable" guess $x_0$, the method will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ In the present case, this will the simply write $$x_{n+1}=\frac{4 x_n^5+10}{5 x_n^4+2}$$ Let us be lazy and start at the middle of the interval  then using $x_0=1.5$. This will generate the following iterates : $1.47826$, $1.47765$ which is the solution for six significant digits.
Edit
You could get a quite accurate approximation using Taylor series built at $x=\frac 32$. Limited to second order, this gives $$x^5+2x-10=\frac{19}{32}+\frac{437}{16} \left(x-\frac{3}{2}\right)+\frac{135}{4}
   \left(x-\frac{3}{2}\right)^2+O\left(\left(x-\frac{3}{2}\right)^3\right)$$ and the solution of the quadratic is $$\frac{1183+\sqrt{170449}}{1080}\approx 1.477643$$ while the exact solution is $\approx 1.477653$.
A: Just for completeness. The real solution of this equation can be expressed in terms of a generalized hypergeometric function
$$
5 \, {}_4F_3\!\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};-\frac{1953125}{512}\right).$$
