How would you find the roots of a function when:
- The function is a polynomial with a degree greater than 3
- The polynomial is not factorable
?
For example (let's just say):
$2x^5-3x^3+13$ I hope this doesn't produce roots with $i$. If you have a clearer example, please add.
As far as I know, I cannot factor this and I have no idea what to do. Is there some alternate form of quadratic formula for higher power polynomials? Is it even possible to do without the use of a computer?
Best Answer
As said in comments, except some very particular cases, there are not explicit expressions for the solutions of quintic polynomials and, most of the time, you will need to use graphics, inspection and numerical methods.
Let us consider the case of $$f(x)=2 x^5-3 x^3+13$$ $$f'(x)=10 x^4-9 x^2$$ $$f''(x)=40 x^3-18 x$$So, the first derivative cancels for $x=0$ (double root) and $x=\pm \frac{3}{\sqrt{10}}$.
So, we have $$f(-\frac{3}{\sqrt{10}})=13+\frac{81}{25 \sqrt{10}}>0\qquad f''(-\frac{3}{\sqrt{10}})=-27 \sqrt{\frac{2}{5}}<0$$ $$f(\frac{3}{\sqrt{10}})=13-\frac{81}{25 \sqrt{10}}>0\qquad f''(\frac{3}{\sqrt{10}})=27 \sqrt{\frac{2}{5}}>0$$ So, the first point is a maximum and second point is a minimum (by the second derivative test) and both of them are positive. So, only one real root is expected.
Using inspection or graphics, you can notice that there is a root between $x=-2$ since $f(-2)=-27$ and $x=-1$ since $f(-1)=14$.
So, now, let us see at a simple numerical method such as Newton for example. Starting from a "reasonable" guess $x_0$, the method will iteratively update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ In the case of the considered function, this will give $$x_{n+1}=\frac{8 x_n^5-6 x_n^3-13}{10x_n^4-9 x_n^2}$$ Let us start at the middle of the interval and the method will generate the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & -1.5000000000000 \\ 1 & -1.76131687242798 \\ 2 & -1.69531261403214 \\ 3 & -1.68843206448835 \\ 4 & -1.68836241590736 \\ 5 & -1.68836240883471 \end{array} \right)$$ which is solution for fifteen significant figures.
The four other roots are complex numbers.