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I looked on Wikipedia for a formula for roots of a 5th degree polynomial, but it said that by Abel's theorem it isn't possible. The Abel's theorem states that you can't solve specific polynomials of the 5th degree using basic operations and root extractions.

Can you find the roots of a specific quintic with only real irrational roots(e.g. $f(x)=x^5+x+2$) using other methods(such as logarithms, trigonometry, or convergent sums of infinite series, etc.)?

Basically, how can the exact values of the roots of such functions be expressed other than a radical(since we know that for some functions it is not a radical)?

If no, is numerical solving/graphing the only way to solve such polynomials?

Edit: I found a link here that explains all the ways that the above mentioned functions could be solved.

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    $\begingroup$ I don't think Abel's theorem states that you can't solve specific polynomials (consider the specific polynomial $(x-1)(x-2)(x-3)(x-4)(x-5)$ for example). Abel's theorem states that there is no general formula (i.e. no analogue of the quadratic formula) that will work for all quintic equations. $\endgroup$ – Sam Weatherhog Dec 2 '15 at 1:00
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    $\begingroup$ You can solve a quintic equation in terms of roots only when it's Galois group is solvable. $\endgroup$ – Sam Weatherhog Dec 2 '15 at 1:01
  • $\begingroup$ @SamWeatherhog there are specific polynomials that cannot be solved in the described way. Of course not every polynomial is such a polynomial, only specific ones. $\endgroup$ – quid Dec 2 '15 at 1:02
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    $\begingroup$ Did you see the section "Beyond radicals" on the WIkipedia page? $\endgroup$ – quid Dec 2 '15 at 1:04
  • $\begingroup$ @quid I think I'm missing your point. Is there an error in what I said? $\endgroup$ – Sam Weatherhog Dec 2 '15 at 2:47
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Here is a summary of posts that should address related questions:

  1. Reducing the general quintic to Bring-Jerrard form.
  2. Solving (1) using elliptic functions.
  3. Reducing the general quintic to Brioschi form.
  4. Solving (3) using trigonometric and special functions.
  5. Solving the Brioschi form using $R(q)$ (or the Rogers-Ramanujan continued fraction).
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As mentioned above, no general formula to find all the roots of any 5th degree equation exists, but various special solution techniques do exist. My own favourite: - By inspection, see if the polynomial has any simple real solutions such as x = 0 or x = 1 or -1 or 2 or -2. If so, divide the poly by (x-a), where a is the found root, and then solve the resultant 4th degree equation by Ferrari's rule. - If no obvious real root exists, one will have to be found. This can be done by noting that if f(p) and f(-p) have different signs, then a root must lie between x=p and x= -p. We now try the halfway point between p and -p, say q. We then repeat the above procedure, continually decreasing the interval in which the root can be found. When the interval is small enough, we have found a root. - This is the bisection method; when such a root has been isolated we divide the polynomial by that root, producing a 4th degree equation which can again be solved by Ferrari or any another method.

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First, if $x=m$ makes a polynomial equal to zero,then $x-m$ is a factor of it,but how to find such $m$ ?

Let's explain the rational root theorem through an example,

consider $$3X^5+x^4-4x^3+8x^2-x+7$$

The theorem says that,if the polynomial has a rational root $\frac{p}{q}\quad(p$ & $q$ are relatively prime$)$, then $p$ is a divisor of the constant term $7$ (negative divisors also), and $q$ is a divisor of the coefficient of $x^n$ (here $n=5$) meaning $3$.

The divisors of $7$ are $1,-1,7,$ and $-7 $, and the divisors of $3$ are $1,-1,3$, and $-3$.

So the possible rational roots are $1,7,\frac{1}{3}, \frac{7}{3}$,and their negatives.

So,you should substitute them into $x$ to see which one makes it zero,then by dividing find the other factors,or you could find ,more than one roots this way,and then do the division.

The substituting can be annoying,for instance how could you find the value of the polynomial for $x=\frac{7}{3}$, but I explained a fast substitution method which is available at the link below.

How about you try above but find no root that way?

In this case,since it is of the degree $5$, then it may be written as $$(ax^3+bx^2+cx+d)(ex^2+fx+g)$$ or $$(ax^4+bx^3+cx^2+dx+e)(fx+g)$$and by doing the multiplication then taking it to be equal to the original polynomial.

Find the unknown coefficients,which can be frustrating.

By checking out the videos in the playlist https://www.youtube.com/playlist?list=PLsMjNmqC7C2T66EgsXhbS-zGiCQB3i2NU, the crazy factoring makes sense to you,and you’ll learn it fast & effortlessly.

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  • $\begingroup$ The last method will not work. After equalling all the coefficients of both polynomials, the resulting equations will still lead you back to the beginning $\endgroup$ – David Jul 15 at 13:40

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