# Finding a polynomial that has solution (root) as the sum of roots

• I
• swampwiz
In summary, finding a polynomial that has a solution (root) as the sum of its roots involves using the Vieta's formulas, which state that the sum of the roots of a polynomial is equal to the negative coefficient of the term with the highest degree divided by the leading coefficient. This can be used to create a polynomial with the desired root as the sum of its roots, by setting the coefficients of the polynomial to the desired values and using the formulas to calculate the remaining coefficient values. This technique is useful in solving polynomial equations and finding the roots of a polynomial with specific conditions.
swampwiz
AIUI, an algebraic is defined as a number that can be the solution (root) of some integer polynomial, and is any number that can be constructed via any binary arithmetic operation or unary root operation with arguments that are themselves algebraic numbers. I have been able to prove this for almost all cases - with the unproven case being where the number is defined as the addition of a pair of algebraic numbers that happen to have a root operation in each.

Here is an example:

x = a1/2 + b1/3

( x - a1/2 ) = b1/3

( x - a1/2 )3 = b

I can't figure out how to continue to de-root a.

The proof of the statement is easier than showing the polynomial. If ##a,b## are algebraic over ##K## then ##b## is algebraic over ##K(a)## and ##a+b\in K(a)(b)=K(a,b)## is algebraic over ##K.##

In order to get the polynomial, write ##(x-a)p(x)=0## and ##(x-b)q(x)=0.## Then ##(x-(a+b)/2)p(x)q(x)=0## and with ##(a+b)/2## algebraic, we have ##a+b## algebraic, too. But you will need the other roots to set up ##p(x)## and ##q(x)##. That's the only way I see.

swampwiz
swampwiz said:
AIUI, an algebraic is defined as a number that can be the solution (root) of some integer polynomial, and is any number that can be constructed via any binary arithmetic operation or unary root operation with arguments that are themselves algebraic numbers. I have been able to prove this for almost all cases - with the unproven case being where the number is defined as the addition of a pair of algebraic numbers that happen to have a root operation in each.

Here is an example:

x = a1/2 + b1/3

( x - a1/2 ) = b1/3

( x - a1/2 )3 = b

I can't figure out how to continue to de-root a.
Expand, solve for root a and square.

##
x^3-3x^2\sqrt{a}+3ax-a\sqrt{a} = b
##

##
\sqrt{a}(3x^2+a) = x^3+3ax-b
##

##
a(3x^2+a)^2 = (x^3+3ax-b)^2
##

I think I was looking at how to do this for the general case of any number of root terms. Only having 2 terms makes it easy to separate, but that can't be done for 3 terms.

x = a1/2 + b1/3 + c1/5

swampwiz said:
I think I was looking at how to do this for the general case of any number of root terms.
Why?

fresh_42 said:
Why?
I'd to prove that any value made of arithmetic & root operations can generate a polynomial that has a solution equal to that value - i.e., this would prove that this type of value is an algebraic number.

swampwiz said:
I'd to prove that any value made of arithmetic & root operations can generate a polynomial that has a solution equal to that value - i.e., this would prove that this type of value is an algebraic number.
This makes no sense. You do not need to know a polynomial for the proof. And without a specific example, you cannot calculate one. You ask for one algorithm that works for any field, for any number of algebraic elements. No way.

fresh_42 said:
This makes no sense. You do not need to know a polynomial for the proof. And without a specific example, you cannot calculate one. You ask for one algorithm that works for any field, for any number of algebraic elements. No way.
If it can be done for any number, then surely there must be some systematic way to do it.

swampwiz said:
If it can be done for any number, then surely there must be some systematic way to do it.
What is any number? Heck, what is even a number? Which fields? Separable extensions? Known Galois group, in case it is Galois? Determine your splitting fields, and calculate all roots. Then apply what I have written in post #2 or proceed as @martinbn suggested in post #3. All you have in the end is Viète.

Your question comes down to: Given any algebraic field extension ##L/K## and an element ##a\in L.## Find its minimal polynomial over ##K##. This means: calculate all powers of ##a## up to ##\dim_K L## and reassemble them with coefficients in ##K## to get ##0.##

That was done in post #3.

Last edited:
fresh_42 said:
The proof of the statement is easier than showing the polynomial. If ##a,b## are algebraic over ##K## then ##b## is algebraic over ##K(a)## and ##a+b\in K(a)(b)=K(a,b)## is algebraic over ##K.##

In order to get the polynomial, write ##(x-a)p(x)=0## and ##(x-b)q(x)=0.## Then ##(x-(a+b)/2)p(x)q(x)=0## and with ##(a+b)/2## algebraic, we have ##a+b## algebraic, too. But you will need the other roots to set up ##p(x)## and ##q(x)##. That's the only way I see.
So what you are saying is:

Because a & b are each algebraic

0 = ( x - a ) = ( x - b )
then for any polynomials p( x ) & q( x )
fresh_42 said:
The proof of the statement is easier than showing the polynomial. If ##a,b## are algebraic over ##K## then ##b## is algebraic over ##K(a)## and ##a+b\in K(a)(b)=K(a,b)## is algebraic over ##K.##

In order to get the polynomial, write ##(x-a)p(x)=0## and ##(x-b)q(x)=0.## Then ##(x-(a+b)/2)p(x)q(x)=0## and with ##(a+b)/2## algebraic, we have ##a+b## algebraic, too. But you will need the other roots to set up ##p(x)## and ##q(x)##. That's the only way I see.
OK, that's a nice little proof that gets me over the hump!

swampwiz said:
So what you are saying is:

Because a & b are each algebraic

0 = ( x - a ) = ( x - b )
?

I mean: because ##\alpha,\beta ## are algebraic, ##K\subseteq K(\alpha ,\beta )## is algebraic. And since ##\alpha +\beta \in K(\alpha ,\beta )## is also algebraic. I would do it in steps: ##K\subseteq K(\alpha )\subseteq K(\alpha ,\beta )##.
swampwiz said:
then for any polynomials p( x ) & q( x )

OK, that's a nice little proof that gets me over the hump!

fresh_42 said:
?

I mean: because ##\alpha,\beta ## are algebraic, ##K\subseteq K(\alpha ,\beta )## is algebraic. And since ##\alpha +\beta \in K(\alpha ,\beta )## is also algebraic. I would do it in steps: ##K\subseteq K(\alpha )\subseteq K(\alpha ,\beta )##.
I had accidentally posted before finishing the post.

0 = ( x - a ) = ( x - b ) = ( x - a ) p( x ) = ( x - b ) q( x ) = ( x - a ) p( x ) q( x ) = ( x - b ) p( x ) q( x )

Summing half of each

0 = ( x - ( a + b ) / 2 ) p( x ) q( x )

thus ( ( a + b ) / 2 ) and as well ( a + b ) must be algebraic.

OK, I think I know what I mean by "an algorithm". If I could prove that any addition, multiplication or root operation having algebraic numbers as arguments results in an algebraic number, then I'm done.

I will now prove why a power, root or product of algebraic numbers is an algebraic number.

GIVEN: a, b -> ALGEBRAIC

THUS: ( a + b ) -> ALGEBRAIC

THUS: ∃ [ f( x ) = 0 = ( x - a ) p( x ) ] &

[ g2( x ) = 0 = ( x + a ) ( x - a ) p( x ) = ( x2 - a2 ) p( x )

[ g3( x ) = 0 = ( x2 + a x + a2 ) ( x - a ) p( x ) = ( x3 - a3 ) p( x )

NOTE: The root operation should be presumed to be a consistent deMoivre root index.

[ g1/2( x ) = 0 = ( x1/2 - a1/2 ) ( x1/2 + a1/2 ) p( x ) = ( x1/2 - a1/2 ) p1/2( x )

[ g1/3( x ) = 0 = ( x1/3 - a1/3 ) ( x2/3 + a1/3 x1/3 + a2/3 ) p( x ) = ( x1/3 - a1/3 ) p1/3( x )

LET: u = xn

THUS: 0 = ( u - an ) p( x( u ) )

THUS: an -> ALGEBRAIC

LET: u = x( 1 / n )

THUS: 0 = ( u - a( 1 / n ) ) p1/n( x( u ) )

THUS: a( 1 / n ) -> ALGEBRAIC

( a b ) = [ ( a + b )2 - ( a2 + b2 ) ] => SUM OF POWERS OF ALGEBRAICS => SUM OF ALGEBRAICS

THUS: ( a b ) -> ALGEBRAIC

QED

Last edited:

## 1. How do you find a polynomial with a given root?

To find a polynomial with a given root, you can use the fact that the sum of the roots of a polynomial is equal to the negative of the coefficient of the second-to-last term divided by the coefficient of the last term. This will give you the value of the missing root, and from there you can use polynomial division to find the rest of the polynomial.

## 2. Can a polynomial have more than one solution (root) that is the sum of the roots?

Yes, a polynomial can have multiple solutions that are the sum of the roots. This is because the sum of the roots is not a unique value and can be achieved by different combinations of roots.

## 3. What is the relationship between the sum of the roots and the coefficient of the second-to-last term in a polynomial?

The sum of the roots of a polynomial is equal to the negative of the coefficient of the second-to-last term divided by the coefficient of the last term. This relationship is known as Vieta's formulas.

## 4. Can you use Vieta's formulas to find the roots of a polynomial?

No, Vieta's formulas only give the sum and product of the roots of a polynomial. To find the actual roots, you will need to use other methods such as factoring or the quadratic formula.

## 5. How can finding a polynomial with a given root be useful in mathematics?

Finding a polynomial with a given root can be useful in various mathematical applications, such as solving equations, graphing polynomials, and understanding the behavior of functions. It can also help in finding the roots of more complex polynomials by using the relationship between the sum of the roots and the coefficients of the polynomial.

• Linear and Abstract Algebra
Replies
3
Views
2K
• Linear and Abstract Algebra
Replies
2
Views
1K
• Math POTW for Secondary and High School Students
Replies
5
Views
1K
• Linear and Abstract Algebra
Replies
15
Views
2K
• Calculus
Replies
4
Views
1K
• Linear and Abstract Algebra
Replies
2
Views
1K
• Linear and Abstract Algebra
Replies
1
Views
1K
• Linear and Abstract Algebra
Replies
6
Views
3K
• Linear and Abstract Algebra
Replies
3
Views
1K
• Linear and Abstract Algebra
Replies
8
Views
2K