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.
functions polynomials roots radicals
 |Â
show 6 more comments
up vote
6
down vote
favorite
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.
functions polynomials roots radicals
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.
â 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.
â 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.
â quidâ¦
Dec 2 '15 at 1:02
1
Did you see the section "Beyond radicals" on the WIkipedia page?
â quidâ¦
Dec 2 '15 at 1:04
@quid I think I'm missing your point. Is there an error in what I said?
â Sam Weatherhog
Dec 2 '15 at 2:47
 |Â
show 6 more comments
up vote
6
down vote
favorite
up vote
6
down vote
favorite
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.
functions polynomials roots radicals
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.
functions polynomials roots radicals
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.
â 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.
â 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.
â quidâ¦
Dec 2 '15 at 1:02
1
Did you see the section "Beyond radicals" on the WIkipedia page?
â quidâ¦
Dec 2 '15 at 1:04
@quid I think I'm missing your point. Is there an error in what I said?
â Sam Weatherhog
Dec 2 '15 at 2:47
 |Â
show 6 more comments
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.
â 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.
â 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.
â quidâ¦
Dec 2 '15 at 1:02
1
Did you see the section "Beyond radicals" on the WIkipedia page?
â quidâ¦
Dec 2 '15 at 1:04
@quid I think I'm missing your point. Is there an error in what I said?
â 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.
â 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.
â 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.
â 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.
â 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.
â quidâ¦
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.
â quidâ¦
Dec 2 '15 at 1:02
1
1
Did you see the section "Beyond radicals" on the WIkipedia page?
â quidâ¦
Dec 2 '15 at 1:04
Did you see the section "Beyond radicals" on the WIkipedia page?
â quidâ¦
Dec 2 '15 at 1:04
@quid I think I'm missing your point. Is there an error in what I said?
â Sam Weatherhog
Dec 2 '15 at 2:47
@quid I think I'm missing your point. Is there an error in what I said?
â Sam Weatherhog
Dec 2 '15 at 2:47
 |Â
show 6 more comments
2 Answers
2
active
oldest
votes
up vote
6
down vote
accepted
Here is a summary of posts that should address related questions:
Reducing the general quintic to Bring-Jerrard form.
Solving (1) using elliptic functions.
Reducing the general quintic to Brioschi form.
Solving (3) using trigonometric and special functions.
Solving the Brioschi form using $R(q)$ (or the Rogers-Ramanujan continued fraction).
add a comment |Â
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.
add a comment |Â
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:
Reducing the general quintic to Bring-Jerrard form.
Solving (1) using elliptic functions.
Reducing the general quintic to Brioschi form.
Solving (3) using trigonometric and special functions.
Solving the Brioschi form using $R(q)$ (or the Rogers-Ramanujan continued fraction).
add a comment |Â
up vote
6
down vote
accepted
Here is a summary of posts that should address related questions:
Reducing the general quintic to Bring-Jerrard form.
Solving (1) using elliptic functions.
Reducing the general quintic to Brioschi form.
Solving (3) using trigonometric and special functions.
Solving the Brioschi form using $R(q)$ (or the Rogers-Ramanujan continued fraction).
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Here is a summary of posts that should address related questions:
Reducing the general quintic to Bring-Jerrard form.
Solving (1) using elliptic functions.
Reducing the general quintic to Brioschi form.
Solving (3) using trigonometric and special functions.
Solving the Brioschi form using $R(q)$ (or the Rogers-Ramanujan continued fraction).
Here is a summary of posts that should address related questions:
Reducing the general quintic to Bring-Jerrard form.
Solving (1) using elliptic functions.
Reducing the general quintic to Brioschi form.
Solving (3) using trigonometric and special functions.
Solving the Brioschi form using $R(q)$ (or the Rogers-Ramanujan continued fraction).
edited Apr 13 '17 at 12:20
Communityâ¦
1
1
answered Dec 3 '15 at 4:26
Tito Piezas III
26.2k361161
26.2k361161
add a comment |Â
add a comment |Â
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.
add a comment |Â
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.
add a comment |Â
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.
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.
answered Jul 19 '17 at 14:18
Peter Bokhout
111
111
add a comment |Â
add a comment |Â
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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.
â 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.
â 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.
â quidâ¦
Dec 2 '15 at 1:02
1
Did you see the section "Beyond radicals" on the WIkipedia page?
â quidâ¦
Dec 2 '15 at 1:04
@quid I think I'm missing your point. Is there an error in what I said?
â Sam Weatherhog
Dec 2 '15 at 2:47