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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?

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

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