How do you solve 5th degree polynomials?

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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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  • 2




    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.
    – Sam Weatherhog
    Dec 2 '15 at 1:00






  • 1




    You can solve a quintic equation in terms of roots only when it's Galois group is solvable.
    – Sam Weatherhog
    Dec 2 '15 at 1:01










  • @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.
    – quid♦
    Dec 2 '15 at 1:02






  • 1




    Did you see the section "Beyond radicals" on the WIkipedia page?
    – quid♦
    Dec 2 '15 at 1:04











  • @quid I think I'm missing your point. Is there an error in what I said?
    – Sam Weatherhog
    Dec 2 '15 at 2:47














up vote
6
down vote

favorite
3












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.







share|cite|improve this question


















  • 2




    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.
    – Sam Weatherhog
    Dec 2 '15 at 1:00






  • 1




    You can solve a quintic equation in terms of roots only when it's Galois group is solvable.
    – Sam Weatherhog
    Dec 2 '15 at 1:01










  • @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.
    – quid♦
    Dec 2 '15 at 1:02






  • 1




    Did you see the section "Beyond radicals" on the WIkipedia page?
    – quid♦
    Dec 2 '15 at 1:04











  • @quid I think I'm missing your point. Is there an error in what I said?
    – Sam Weatherhog
    Dec 2 '15 at 2:47












up vote
6
down vote

favorite
3









up vote
6
down vote

favorite
3






3





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.







share|cite|improve this question














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.









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '15 at 1:43

























asked Dec 2 '15 at 0:55









Stepanich02

11216




11216







  • 2




    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.
    – Sam Weatherhog
    Dec 2 '15 at 1:00






  • 1




    You can solve a quintic equation in terms of roots only when it's Galois group is solvable.
    – Sam Weatherhog
    Dec 2 '15 at 1:01










  • @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.
    – quid♦
    Dec 2 '15 at 1:02






  • 1




    Did you see the section "Beyond radicals" on the WIkipedia page?
    – quid♦
    Dec 2 '15 at 1:04











  • @quid I think I'm missing your point. Is there an error in what I said?
    – Sam Weatherhog
    Dec 2 '15 at 2:47












  • 2




    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.
    – Sam Weatherhog
    Dec 2 '15 at 1:00






  • 1




    You can solve a quintic equation in terms of roots only when it's Galois group is solvable.
    – Sam Weatherhog
    Dec 2 '15 at 1:01










  • @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.
    – quid♦
    Dec 2 '15 at 1:02






  • 1




    Did you see the section "Beyond radicals" on the WIkipedia page?
    – quid♦
    Dec 2 '15 at 1:04











  • @quid I think I'm missing your point. Is there an error in what I said?
    – Sam Weatherhog
    Dec 2 '15 at 2:47







2




2




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.
– Sam Weatherhog
Dec 2 '15 at 1:00




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.
– Sam Weatherhog
Dec 2 '15 at 1:00




1




1




You can solve a quintic equation in terms of roots only when it's Galois group is solvable.
– Sam Weatherhog
Dec 2 '15 at 1:01




You can solve a quintic equation in terms of roots only when it's Galois group is solvable.
– Sam Weatherhog
Dec 2 '15 at 1:01












@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.
– quid♦
Dec 2 '15 at 1:02




@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.
– quid♦
Dec 2 '15 at 1:02




1




1




Did you see the section "Beyond radicals" on the WIkipedia page?
– quid♦
Dec 2 '15 at 1:04





Did you see the section "Beyond radicals" on the WIkipedia page?
– quid♦
Dec 2 '15 at 1:04













@quid I think I'm missing your point. Is there an error in what I said?
– Sam Weatherhog
Dec 2 '15 at 2:47




@quid I think I'm missing your point. Is there an error in what I said?
– Sam Weatherhog
Dec 2 '15 at 2:47










2 Answers
2






active

oldest

votes

















up vote
6
down vote



accepted










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).





share|cite|improve this answer





























    up vote
    1
    down vote













    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.






    share|cite|improve this answer




















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      6
      down vote



      accepted










      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).





      share|cite|improve this answer


























        up vote
        6
        down vote



        accepted










        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).





        share|cite|improve this answer
























          up vote
          6
          down vote



          accepted







          up vote
          6
          down vote



          accepted






          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).





          share|cite|improve this answer














          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).






          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Apr 13 '17 at 12:20









          Community♦

          1




          1










          answered Dec 3 '15 at 4:26









          Tito Piezas III

          26.2k361161




          26.2k361161




















              up vote
              1
              down vote













              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.






              share|cite|improve this answer
























                up vote
                1
                down vote













                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.






                share|cite|improve this answer






















                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  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.






                  share|cite|improve this answer












                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jul 19 '17 at 14:18









                  Peter Bokhout

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