# How to find the roots of a higher degree polynomial?

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?

• You won't necessarily be able to. There are polynomials of degree $\geq 5$ with real roots which cannot be expressed in terms of radicals. – MathematicsStudent1122 Oct 14 '16 at 3:48
• 5 and above cannot be done. You need to study group theory to know why. – user2277550 Oct 14 '16 at 3:49
• Ok, I'm only in calc 1 so I guess it will be a while. – chef stu Oct 14 '16 at 3:51
• @MathematicsStudent1122 there is some solvable $5th$ degree equations which can be solved by some formulas such as De moivre's quantic. – Pentapolis Oct 14 '16 at 3:52
• @Pentapolis I never said there weren't. "There are polynomials...." $\neq$ "All polynomials....". – MathematicsStudent1122 Oct 14 '16 at 3:56

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.