What does it mean to factor over the real numbers?
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I am confused on the topic of factoring over real numbers. What is the difference between normally factoring and factoring over real numbers? If anyone could explain, that would be appreciated! Thanks ahead of time!
factoring
asked Jan 5 at 22:30
JamesJames
264
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add a comment |
$begingroup$
I am confused on the topic of factoring over real numbers. What is the difference between normally factoring and factoring over real numbers? If anyone could explain, that would be appreciated! Thanks ahead of time!
factoring
asked Jan 5 at 22:30
JamesJames
264
$endgroup$
$begingroup$
If you're asking about factoring polynomials then "factoring ovr the real numbers" means that the factors should be polynomials all of whose coefficients are real. If you're asking about something else, then you should say what you intend.
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– Andreas Blass
Jan 5 at 22:38
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Yes, I am asking about factoring polynomials. For instance, could you explain to me how to factor the following polynomial: 2x^4-5x^3-4x^2+15x-6
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– James
Jan 5 at 22:41
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Could you put "factor over the reals" in context? And what are you factoring? polynomials? numbers? It seems to me that the difference between of reals and integers is that you have any two reals $a,b$ and $b ne 0$ then you can always find a real $k$ so that $a = b*k$ so every non-zero number is a factor of every number whereas under integers to have $b|a$ is a special event that may not occur.
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– fleablood
Jan 5 at 22:42
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I am speaking in terms of polynomials.
$endgroup$
– James
Jan 5 at 22:48
add a comment |
$begingroup$
I am confused on the topic of factoring over real numbers. What is the difference between normally factoring and factoring over real numbers? If anyone could explain, that would be appreciated! Thanks ahead of time!
factoring
asked Jan 5 at 22:30
JamesJames
264
$endgroup$
I am confused on the topic of factoring over real numbers. What is the difference between normally factoring and factoring over real numbers? If anyone could explain, that would be appreciated! Thanks ahead of time!
factoring
factoring
asked Jan 5 at 22:30
JamesJames
264
asked Jan 5 at 22:30
JamesJames
264
edited Jan 5 at 22:36
James
asked Jan 5 at 22:30
JamesJames
264
asked Jan 5 at 22:30
JamesJames
264
asked Jan 5 at 22:30
JamesJames
264
264
$begingroup$
If you're asking about factoring polynomials then "factoring ovr the real numbers" means that the factors should be polynomials all of whose coefficients are real. If you're asking about something else, then you should say what you intend.
$endgroup$
– Andreas Blass
Jan 5 at 22:38
$begingroup$
Yes, I am asking about factoring polynomials. For instance, could you explain to me how to factor the following polynomial: 2x^4-5x^3-4x^2+15x-6
$endgroup$
– James
Jan 5 at 22:41
$begingroup$
Could you put "factor over the reals" in context? And what are you factoring? polynomials? numbers? It seems to me that the difference between of reals and integers is that you have any two reals $a,b$ and $b ne 0$ then you can always find a real $k$ so that $a = b*k$ so every non-zero number is a factor of every number whereas under integers to have $b|a$ is a special event that may not occur.
$endgroup$
– fleablood
Jan 5 at 22:42
$begingroup$
I am speaking in terms of polynomials.
$endgroup$
– James
Jan 5 at 22:48
add a comment |
$begingroup$
If you're asking about factoring polynomials then "factoring ovr the real numbers" means that the factors should be polynomials all of whose coefficients are real. If you're asking about something else, then you should say what you intend.
$endgroup$
– Andreas Blass
Jan 5 at 22:38
$begingroup$
Yes, I am asking about factoring polynomials. For instance, could you explain to me how to factor the following polynomial: 2x^4-5x^3-4x^2+15x-6
$endgroup$
– James
Jan 5 at 22:41
$begingroup$
Could you put "factor over the reals" in context? And what are you factoring? polynomials? numbers? It seems to me that the difference between of reals and integers is that you have any two reals $a,b$ and $b ne 0$ then you can always find a real $k$ so that $a = b*k$ so every non-zero number is a factor of every number whereas under integers to have $b|a$ is a special event that may not occur.
$endgroup$
– fleablood
Jan 5 at 22:42
$begingroup$
I am speaking in terms of polynomials.
$endgroup$
– James
Jan 5 at 22:48
$begingroup$
If you're asking about factoring polynomials then "factoring ovr the real numbers" means that the factors should be polynomials all of whose coefficients are real. If you're asking about something else, then you should say what you intend.
$endgroup$
– Andreas Blass
Jan 5 at 22:38
$begingroup$
If you're asking about factoring polynomials then "factoring ovr the real numbers" means that the factors should be polynomials all of whose coefficients are real. If you're asking about something else, then you should say what you intend.
$endgroup$
– Andreas Blass
Jan 5 at 22:38
$begingroup$
Yes, I am asking about factoring polynomials. For instance, could you explain to me how to factor the following polynomial: 2x^4-5x^3-4x^2+15x-6
$endgroup$
– James
Jan 5 at 22:41
$begingroup$
Yes, I am asking about factoring polynomials. For instance, could you explain to me how to factor the following polynomial: 2x^4-5x^3-4x^2+15x-6
$endgroup$
– James
Jan 5 at 22:41
$begingroup$
Could you put "factor over the reals" in context? And what are you factoring? polynomials? numbers? It seems to me that the difference between of reals and integers is that you have any two reals $a,b$ and $b ne 0$ then you can always find a real $k$ so that $a = b*k$ so every non-zero number is a factor of every number whereas under integers to have $b|a$ is a special event that may not occur.
$endgroup$
– fleablood
Jan 5 at 22:42
$begingroup$
Could you put "factor over the reals" in context? And what are you factoring? polynomials? numbers? It seems to me that the difference between of reals and integers is that you have any two reals $a,b$ and $b ne 0$ then you can always find a real $k$ so that $a = b*k$ so every non-zero number is a factor of every number whereas under integers to have $b|a$ is a special event that may not occur.
$endgroup$
– fleablood
Jan 5 at 22:42
$begingroup$
I am speaking in terms of polynomials.
$endgroup$
– James
Jan 5 at 22:48
$begingroup$
I am speaking in terms of polynomials.
$endgroup$
– James
Jan 5 at 22:48
add a comment |
3 Answers
3
active
oldest
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Factoring, as one learns in elementary algebra and high school, is always done “over the real numbers”. What this means is that when we factor a polynomial, the factors should be in the reals.
Later on, we become interested in factoring over other “fields”. An example of this is say we ask if $x^2-2$ is factorable over the rationals. This can be written as $(x-sqrt2)(x+sqrt2)$, by difference of squares. Note that $sqrt2$ is not a rational number, so this polynomial is not factorable over the rationals. (Note however it is over the reals).
Not sure if you have been exposed to this, but as an interesting idea, imagine we invented a number $i$ with $i^2=-1$. This “$i$” is clearly not a real number, so we can imagine a new number system of the form $a+bi$, where $a$,$b$ are real. We’ll call this the complex numbers. With this in mind, let us ask if $x^2+1$ is factorable. Over the real numbers, you will likely have trouble factoring this. However, over the complex numbers we just defined, we can write this as $(x+i)(x-i)$, meaning that $x^2+1$ is not factorable over the real numbers but it is over the complex numbers!
Hopefully this demonstrates the importance of the field you are factoring over.
answered Jan 5 at 22:49
Tyler6Tyler6
838413
$endgroup$
1
$begingroup$
"Factoring, as one learns in elementary algebra and high school," My millage varied. In my high school we learned to factor over the rationals and not the reals. We would not have been been taught $x^2 - 2$ can not be "factored".
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– fleablood
Jan 5 at 22:55
add a comment |
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Not all polynomials have rational roots. For an easy example, take the polynomial $x^2-2=0$. Factoring over the rationals, there are no answers. Factoring over the reals, there are two answers, $pm sqrt2$.
answered Jan 5 at 22:41
BadAtGeometryBadAtGeometry
188215
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So could 2^4-5x^3-4x^2+15x-6 be factored over real numbers? Would the answer to the polynomial be (x-2)(2x-1)(x^2-3) or (x-2)(2x-1)(x-root3)(x-root3)?
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– James
Jan 5 at 22:47
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@MichaelBoulis If you want to factor over the rationals, it would be the first case-(x-2)(2x-1)(x^2-3) But if you want to factor over the reals, it would be the second case-(x-2)(2x-1)(x-root3)(x-root3).
$endgroup$
– BadAtGeometry
Jan 5 at 22:49
$begingroup$
So what really limits factoring over reals, as in what cannot be factored over real numbers since, as taught in middle and high school mathematics, real numbers are the domain which consist of every possible number.
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– James
Jan 5 at 22:53
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Yes, but if you have learned about complex numbers, check out the second part of @Tyler6s post.
$endgroup$
– BadAtGeometry
Jan 5 at 22:55
add a comment |
$begingroup$
A simple example would be factoring $x^2 - 2=0$.
This polynomial has no rational roots so it can not be factored over the rationals.
But it has two real roots $sqrt{2}$ and $-sqrt{2}$ so if we factor it over the reals then it factors as $x^2 -2 = (x - sqrt{2})(x + sqrt{2})$.
A more illustrative example might be $x^3 - 2x^2 - x + 2$.
By the rational roots test and trial and error it has a root of $1$ ($1^3 -2 - 1 + 2 = 0$) so we can factor $x-1$ out of it and get:
$x^2 - 2x^2 - x + 2 = (x-1)(x^2 - 2)$ and $x^2 - 2$ can't be factored over the rationals so that is our final factorization.
But over the reals $x^2-2$ has two roots so it may be factored as $(x-1)(x+sqrt 2)(x - sqrt 2)$.
Finally consider $x^2 + 2 = 0$. This has no real roots at all so it can not be factored of the reals. (Nor over the rationals.)
And $x^3 - x^2 +2x - 2$ has only one real root. $1$. Which is a rational root. It can only be factored as far as $(x -1)(x^2 + 2)$ and that's as far as you can factor it over the rationals or over the reals.
answered Jan 5 at 22:53
fleabloodfleablood
72k22687
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On Michael Wang's comment concerning the factoring over the polynomial 2x^4-5x^3-4x^2+15x-6, I checked over his comment and found that his factoring the polynomial over the real numbers is incorrect. What would the correct solution be (add explanation if possible)?
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– James
Jan 5 at 23:16
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I believe the correct solution to the complete factoring over real numbers of 2x^4-5x^3-4x^2+15x-6 is (x-2)(2x-1)(x-root3)(x+root3). Correct me if I am wrong.
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– James
Jan 5 at 23:22
$begingroup$
If M.Wangs answer has an error its based on your work but it has no error. $x=2$ is a solution and we can factor $2x^4 - 5x^3 -4x^2+15x-6= 2x^4 - 4x^3 - x^3 +2x^2 -6x^2+12x + 3x -6 = (x-2)(2x^3-x^2 -6x+3)$ and we can factor $2x-1$ from that to get $(x-2)(2x-1)(2x-1)(x^2-3)$ and .... as you should be becoming familiar with $x^2 -3$ can not be factored over the rationals but it can be factor as $(x-sqrt3)(x+sqrt 3)$ so $(x-2)(2x-1)(2x-1)(x^2-3)$ is complete factorizing over the rationals and $(x-2)(2x-1)(2x-1)(x-sqrt3)(x+sqrt 3)$ is complete factorization over reals.
$endgroup$
– fleablood
Jan 6 at 4:08
add a comment |
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3 Answers
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$begingroup$
Factoring, as one learns in elementary algebra and high school, is always done “over the real numbers”. What this means is that when we factor a polynomial, the factors should be in the reals.
Later on, we become interested in factoring over other “fields”. An example of this is say we ask if $x^2-2$ is factorable over the rationals. This can be written as $(x-sqrt2)(x+sqrt2)$, by difference of squares. Note that $sqrt2$ is not a rational number, so this polynomial is not factorable over the rationals. (Note however it is over the reals).
Not sure if you have been exposed to this, but as an interesting idea, imagine we invented a number $i$ with $i^2=-1$. This “$i$” is clearly not a real number, so we can imagine a new number system of the form $a+bi$, where $a$,$b$ are real. We’ll call this the complex numbers. With this in mind, let us ask if $x^2+1$ is factorable. Over the real numbers, you will likely have trouble factoring this. However, over the complex numbers we just defined, we can write this as $(x+i)(x-i)$, meaning that $x^2+1$ is not factorable over the real numbers but it is over the complex numbers!
Hopefully this demonstrates the importance of the field you are factoring over.
answered Jan 5 at 22:49
Tyler6Tyler6
838413
$endgroup$
1
$begingroup$
"Factoring, as one learns in elementary algebra and high school," My millage varied. In my high school we learned to factor over the rationals and not the reals. We would not have been been taught $x^2 - 2$ can not be "factored".
$endgroup$
– fleablood
Jan 5 at 22:55
add a comment |
$begingroup$
Factoring, as one learns in elementary algebra and high school, is always done “over the real numbers”. What this means is that when we factor a polynomial, the factors should be in the reals.
Later on, we become interested in factoring over other “fields”. An example of this is say we ask if $x^2-2$ is factorable over the rationals. This can be written as $(x-sqrt2)(x+sqrt2)$, by difference of squares. Note that $sqrt2$ is not a rational number, so this polynomial is not factorable over the rationals. (Note however it is over the reals).
Not sure if you have been exposed to this, but as an interesting idea, imagine we invented a number $i$ with $i^2=-1$. This “$i$” is clearly not a real number, so we can imagine a new number system of the form $a+bi$, where $a$,$b$ are real. We’ll call this the complex numbers. With this in mind, let us ask if $x^2+1$ is factorable. Over the real numbers, you will likely have trouble factoring this. However, over the complex numbers we just defined, we can write this as $(x+i)(x-i)$, meaning that $x^2+1$ is not factorable over the real numbers but it is over the complex numbers!
Hopefully this demonstrates the importance of the field you are factoring over.
answered Jan 5 at 22:49
Tyler6Tyler6
838413
$endgroup$
1
$begingroup$
"Factoring, as one learns in elementary algebra and high school," My millage varied. In my high school we learned to factor over the rationals and not the reals. We would not have been been taught $x^2 - 2$ can not be "factored".
$endgroup$
– fleablood
Jan 5 at 22:55
add a comment |
$begingroup$
Factoring, as one learns in elementary algebra and high school, is always done “over the real numbers”. What this means is that when we factor a polynomial, the factors should be in the reals.
Later on, we become interested in factoring over other “fields”. An example of this is say we ask if $x^2-2$ is factorable over the rationals. This can be written as $(x-sqrt2)(x+sqrt2)$, by difference of squares. Note that $sqrt2$ is not a rational number, so this polynomial is not factorable over the rationals. (Note however it is over the reals).
Not sure if you have been exposed to this, but as an interesting idea, imagine we invented a number $i$ with $i^2=-1$. This “$i$” is clearly not a real number, so we can imagine a new number system of the form $a+bi$, where $a$,$b$ are real. We’ll call this the complex numbers. With this in mind, let us ask if $x^2+1$ is factorable. Over the real numbers, you will likely have trouble factoring this. However, over the complex numbers we just defined, we can write this as $(x+i)(x-i)$, meaning that $x^2+1$ is not factorable over the real numbers but it is over the complex numbers!
Hopefully this demonstrates the importance of the field you are factoring over.
answered Jan 5 at 22:49
Tyler6Tyler6
838413
$endgroup$
Factoring, as one learns in elementary algebra and high school, is always done “over the real numbers”. What this means is that when we factor a polynomial, the factors should be in the reals.
Later on, we become interested in factoring over other “fields”. An example of this is say we ask if $x^2-2$ is factorable over the rationals. This can be written as $(x-sqrt2)(x+sqrt2)$, by difference of squares. Note that $sqrt2$ is not a rational number, so this polynomial is not factorable over the rationals. (Note however it is over the reals).
Not sure if you have been exposed to this, but as an interesting idea, imagine we invented a number $i$ with $i^2=-1$. This “$i$” is clearly not a real number, so we can imagine a new number system of the form $a+bi$, where $a$,$b$ are real. We’ll call this the complex numbers. With this in mind, let us ask if $x^2+1$ is factorable. Over the real numbers, you will likely have trouble factoring this. However, over the complex numbers we just defined, we can write this as $(x+i)(x-i)$, meaning that $x^2+1$ is not factorable over the real numbers but it is over the complex numbers!
Hopefully this demonstrates the importance of the field you are factoring over.
answered Jan 5 at 22:49
Tyler6Tyler6
838413
answered Jan 5 at 22:49
Tyler6Tyler6
838413
answered Jan 5 at 22:49
Tyler6Tyler6
838413
answered Jan 5 at 22:49
Tyler6Tyler6
838413
838413
1
$begingroup$
"Factoring, as one learns in elementary algebra and high school," My millage varied. In my high school we learned to factor over the rationals and not the reals. We would not have been been taught $x^2 - 2$ can not be "factored".
$endgroup$
– fleablood
Jan 5 at 22:55
add a comment |
1
$begingroup$
"Factoring, as one learns in elementary algebra and high school," My millage varied. In my high school we learned to factor over the rationals and not the reals. We would not have been been taught $x^2 - 2$ can not be "factored".
$endgroup$
– fleablood
Jan 5 at 22:55
1
1
$begingroup$
"Factoring, as one learns in elementary algebra and high school," My millage varied. In my high school we learned to factor over the rationals and not the reals. We would not have been been taught $x^2 - 2$ can not be "factored".
$endgroup$
– fleablood
Jan 5 at 22:55
$begingroup$
"Factoring, as one learns in elementary algebra and high school," My millage varied. In my high school we learned to factor over the rationals and not the reals. We would not have been been taught $x^2 - 2$ can not be "factored".
$endgroup$
– fleablood
Jan 5 at 22:55
add a comment |
$begingroup$
Not all polynomials have rational roots. For an easy example, take the polynomial $x^2-2=0$. Factoring over the rationals, there are no answers. Factoring over the reals, there are two answers, $pm sqrt2$.
answered Jan 5 at 22:41
BadAtGeometryBadAtGeometry
188215
$endgroup$
$begingroup$
So could 2^4-5x^3-4x^2+15x-6 be factored over real numbers? Would the answer to the polynomial be (x-2)(2x-1)(x^2-3) or (x-2)(2x-1)(x-root3)(x-root3)?
$endgroup$
– James
Jan 5 at 22:47
$begingroup$
@MichaelBoulis If you want to factor over the rationals, it would be the first case-(x-2)(2x-1)(x^2-3) But if you want to factor over the reals, it would be the second case-(x-2)(2x-1)(x-root3)(x-root3).
$endgroup$
– BadAtGeometry
Jan 5 at 22:49
$begingroup$
So what really limits factoring over reals, as in what cannot be factored over real numbers since, as taught in middle and high school mathematics, real numbers are the domain which consist of every possible number.
$endgroup$
– James
Jan 5 at 22:53
$begingroup$
Yes, but if you have learned about complex numbers, check out the second part of @Tyler6s post.
$endgroup$
– BadAtGeometry
Jan 5 at 22:55
add a comment |
$begingroup$
Not all polynomials have rational roots. For an easy example, take the polynomial $x^2-2=0$. Factoring over the rationals, there are no answers. Factoring over the reals, there are two answers, $pm sqrt2$.
answered Jan 5 at 22:41
BadAtGeometryBadAtGeometry
188215
$endgroup$
$begingroup$
So could 2^4-5x^3-4x^2+15x-6 be factored over real numbers? Would the answer to the polynomial be (x-2)(2x-1)(x^2-3) or (x-2)(2x-1)(x-root3)(x-root3)?
$endgroup$
– James
Jan 5 at 22:47
$begingroup$
@MichaelBoulis If you want to factor over the rationals, it would be the first case-(x-2)(2x-1)(x^2-3) But if you want to factor over the reals, it would be the second case-(x-2)(2x-1)(x-root3)(x-root3).
$endgroup$
– BadAtGeometry
Jan 5 at 22:49
$begingroup$
So what really limits factoring over reals, as in what cannot be factored over real numbers since, as taught in middle and high school mathematics, real numbers are the domain which consist of every possible number.
$endgroup$
– James
Jan 5 at 22:53
$begingroup$
Yes, but if you have learned about complex numbers, check out the second part of @Tyler6s post.
$endgroup$
– BadAtGeometry
Jan 5 at 22:55
add a comment |
$begingroup$
Not all polynomials have rational roots. For an easy example, take the polynomial $x^2-2=0$. Factoring over the rationals, there are no answers. Factoring over the reals, there are two answers, $pm sqrt2$.
answered Jan 5 at 22:41
BadAtGeometryBadAtGeometry
188215
$endgroup$
Not all polynomials have rational roots. For an easy example, take the polynomial $x^2-2=0$. Factoring over the rationals, there are no answers. Factoring over the reals, there are two answers, $pm sqrt2$.
answered Jan 5 at 22:41
BadAtGeometryBadAtGeometry
188215
answered Jan 5 at 22:41
BadAtGeometryBadAtGeometry
188215
answered Jan 5 at 22:41
BadAtGeometryBadAtGeometry
188215
answered Jan 5 at 22:41
BadAtGeometryBadAtGeometry
188215
188215
$begingroup$
So could 2^4-5x^3-4x^2+15x-6 be factored over real numbers? Would the answer to the polynomial be (x-2)(2x-1)(x^2-3) or (x-2)(2x-1)(x-root3)(x-root3)?
$endgroup$
– James
Jan 5 at 22:47
$begingroup$
@MichaelBoulis If you want to factor over the rationals, it would be the first case-(x-2)(2x-1)(x^2-3) But if you want to factor over the reals, it would be the second case-(x-2)(2x-1)(x-root3)(x-root3).
$endgroup$
– BadAtGeometry
Jan 5 at 22:49
$begingroup$
So what really limits factoring over reals, as in what cannot be factored over real numbers since, as taught in middle and high school mathematics, real numbers are the domain which consist of every possible number.
$endgroup$
– James
Jan 5 at 22:53
$begingroup$
Yes, but if you have learned about complex numbers, check out the second part of @Tyler6s post.
$endgroup$
– BadAtGeometry
Jan 5 at 22:55
add a comment |
$begingroup$
So could 2^4-5x^3-4x^2+15x-6 be factored over real numbers? Would the answer to the polynomial be (x-2)(2x-1)(x^2-3) or (x-2)(2x-1)(x-root3)(x-root3)?
$endgroup$
– James
Jan 5 at 22:47
$begingroup$
@MichaelBoulis If you want to factor over the rationals, it would be the first case-(x-2)(2x-1)(x^2-3) But if you want to factor over the reals, it would be the second case-(x-2)(2x-1)(x-root3)(x-root3).
$endgroup$
– BadAtGeometry
Jan 5 at 22:49
$begingroup$
So what really limits factoring over reals, as in what cannot be factored over real numbers since, as taught in middle and high school mathematics, real numbers are the domain which consist of every possible number.
$endgroup$
– James
Jan 5 at 22:53
$begingroup$
Yes, but if you have learned about complex numbers, check out the second part of @Tyler6s post.
$endgroup$
– BadAtGeometry
Jan 5 at 22:55
$begingroup$
So could 2^4-5x^3-4x^2+15x-6 be factored over real numbers? Would the answer to the polynomial be (x-2)(2x-1)(x^2-3) or (x-2)(2x-1)(x-root3)(x-root3)?
$endgroup$
– James
Jan 5 at 22:47
$begingroup$
So could 2^4-5x^3-4x^2+15x-6 be factored over real numbers? Would the answer to the polynomial be (x-2)(2x-1)(x^2-3) or (x-2)(2x-1)(x-root3)(x-root3)?
$endgroup$
– James
Jan 5 at 22:47
$begingroup$
@MichaelBoulis If you want to factor over the rationals, it would be the first case-(x-2)(2x-1)(x^2-3) But if you want to factor over the reals, it would be the second case-(x-2)(2x-1)(x-root3)(x-root3).
$endgroup$
– BadAtGeometry
Jan 5 at 22:49
$begingroup$
@MichaelBoulis If you want to factor over the rationals, it would be the first case-(x-2)(2x-1)(x^2-3) But if you want to factor over the reals, it would be the second case-(x-2)(2x-1)(x-root3)(x-root3).
$endgroup$
– BadAtGeometry
Jan 5 at 22:49
$begingroup$
So what really limits factoring over reals, as in what cannot be factored over real numbers since, as taught in middle and high school mathematics, real numbers are the domain which consist of every possible number.
$endgroup$
– James
Jan 5 at 22:53
$begingroup$
So what really limits factoring over reals, as in what cannot be factored over real numbers since, as taught in middle and high school mathematics, real numbers are the domain which consist of every possible number.
$endgroup$
– James
Jan 5 at 22:53
$begingroup$
Yes, but if you have learned about complex numbers, check out the second part of @Tyler6s post.
$endgroup$
– BadAtGeometry
Jan 5 at 22:55
$begingroup$
Yes, but if you have learned about complex numbers, check out the second part of @Tyler6s post.
$endgroup$
– BadAtGeometry
Jan 5 at 22:55
add a comment |
$begingroup$
A simple example would be factoring $x^2 - 2=0$.
This polynomial has no rational roots so it can not be factored over the rationals.
But it has two real roots $sqrt{2}$ and $-sqrt{2}$ so if we factor it over the reals then it factors as $x^2 -2 = (x - sqrt{2})(x + sqrt{2})$.
A more illustrative example might be $x^3 - 2x^2 - x + 2$.
By the rational roots test and trial and error it has a root of $1$ ($1^3 -2 - 1 + 2 = 0$) so we can factor $x-1$ out of it and get:
$x^2 - 2x^2 - x + 2 = (x-1)(x^2 - 2)$ and $x^2 - 2$ can't be factored over the rationals so that is our final factorization.
But over the reals $x^2-2$ has two roots so it may be factored as $(x-1)(x+sqrt 2)(x - sqrt 2)$.
Finally consider $x^2 + 2 = 0$. This has no real roots at all so it can not be factored of the reals. (Nor over the rationals.)
And $x^3 - x^2 +2x - 2$ has only one real root. $1$. Which is a rational root. It can only be factored as far as $(x -1)(x^2 + 2)$ and that's as far as you can factor it over the rationals or over the reals.
answered Jan 5 at 22:53
fleabloodfleablood
72k22687
$endgroup$
$begingroup$
On Michael Wang's comment concerning the factoring over the polynomial 2x^4-5x^3-4x^2+15x-6, I checked over his comment and found that his factoring the polynomial over the real numbers is incorrect. What would the correct solution be (add explanation if possible)?
$endgroup$
– James
Jan 5 at 23:16
$begingroup$
I believe the correct solution to the complete factoring over real numbers of 2x^4-5x^3-4x^2+15x-6 is (x-2)(2x-1)(x-root3)(x+root3). Correct me if I am wrong.
$endgroup$
– James
Jan 5 at 23:22
$begingroup$
If M.Wangs answer has an error its based on your work but it has no error. $x=2$ is a solution and we can factor $2x^4 - 5x^3 -4x^2+15x-6= 2x^4 - 4x^3 - x^3 +2x^2 -6x^2+12x + 3x -6 = (x-2)(2x^3-x^2 -6x+3)$ and we can factor $2x-1$ from that to get $(x-2)(2x-1)(2x-1)(x^2-3)$ and .... as you should be becoming familiar with $x^2 -3$ can not be factored over the rationals but it can be factor as $(x-sqrt3)(x+sqrt 3)$ so $(x-2)(2x-1)(2x-1)(x^2-3)$ is complete factorizing over the rationals and $(x-2)(2x-1)(2x-1)(x-sqrt3)(x+sqrt 3)$ is complete factorization over reals.
$endgroup$
– fleablood
Jan 6 at 4:08
add a comment |
$begingroup$
A simple example would be factoring $x^2 - 2=0$.
This polynomial has no rational roots so it can not be factored over the rationals.
But it has two real roots $sqrt{2}$ and $-sqrt{2}$ so if we factor it over the reals then it factors as $x^2 -2 = (x - sqrt{2})(x + sqrt{2})$.
A more illustrative example might be $x^3 - 2x^2 - x + 2$.
By the rational roots test and trial and error it has a root of $1$ ($1^3 -2 - 1 + 2 = 0$) so we can factor $x-1$ out of it and get:
$x^2 - 2x^2 - x + 2 = (x-1)(x^2 - 2)$ and $x^2 - 2$ can't be factored over the rationals so that is our final factorization.
But over the reals $x^2-2$ has two roots so it may be factored as $(x-1)(x+sqrt 2)(x - sqrt 2)$.
Finally consider $x^2 + 2 = 0$. This has no real roots at all so it can not be factored of the reals. (Nor over the rationals.)
And $x^3 - x^2 +2x - 2$ has only one real root. $1$. Which is a rational root. It can only be factored as far as $(x -1)(x^2 + 2)$ and that's as far as you can factor it over the rationals or over the reals.
answered Jan 5 at 22:53
fleabloodfleablood
72k22687
$endgroup$
$begingroup$
On Michael Wang's comment concerning the factoring over the polynomial 2x^4-5x^3-4x^2+15x-6, I checked over his comment and found that his factoring the polynomial over the real numbers is incorrect. What would the correct solution be (add explanation if possible)?
$endgroup$
– James
Jan 5 at 23:16
$begingroup$
I believe the correct solution to the complete factoring over real numbers of 2x^4-5x^3-4x^2+15x-6 is (x-2)(2x-1)(x-root3)(x+root3). Correct me if I am wrong.
$endgroup$
– James
Jan 5 at 23:22
$begingroup$
If M.Wangs answer has an error its based on your work but it has no error. $x=2$ is a solution and we can factor $2x^4 - 5x^3 -4x^2+15x-6= 2x^4 - 4x^3 - x^3 +2x^2 -6x^2+12x + 3x -6 = (x-2)(2x^3-x^2 -6x+3)$ and we can factor $2x-1$ from that to get $(x-2)(2x-1)(2x-1)(x^2-3)$ and .... as you should be becoming familiar with $x^2 -3$ can not be factored over the rationals but it can be factor as $(x-sqrt3)(x+sqrt 3)$ so $(x-2)(2x-1)(2x-1)(x^2-3)$ is complete factorizing over the rationals and $(x-2)(2x-1)(2x-1)(x-sqrt3)(x+sqrt 3)$ is complete factorization over reals.
$endgroup$
– fleablood
Jan 6 at 4:08
add a comment |
$begingroup$
A simple example would be factoring $x^2 - 2=0$.
This polynomial has no rational roots so it can not be factored over the rationals.
But it has two real roots $sqrt{2}$ and $-sqrt{2}$ so if we factor it over the reals then it factors as $x^2 -2 = (x - sqrt{2})(x + sqrt{2})$.
A more illustrative example might be $x^3 - 2x^2 - x + 2$.
By the rational roots test and trial and error it has a root of $1$ ($1^3 -2 - 1 + 2 = 0$) so we can factor $x-1$ out of it and get:
$x^2 - 2x^2 - x + 2 = (x-1)(x^2 - 2)$ and $x^2 - 2$ can't be factored over the rationals so that is our final factorization.
But over the reals $x^2-2$ has two roots so it may be factored as $(x-1)(x+sqrt 2)(x - sqrt 2)$.
Finally consider $x^2 + 2 = 0$. This has no real roots at all so it can not be factored of the reals. (Nor over the rationals.)
And $x^3 - x^2 +2x - 2$ has only one real root. $1$. Which is a rational root. It can only be factored as far as $(x -1)(x^2 + 2)$ and that's as far as you can factor it over the rationals or over the reals.
answered Jan 5 at 22:53
fleabloodfleablood
72k22687
$endgroup$
A simple example would be factoring $x^2 - 2=0$.
This polynomial has no rational roots so it can not be factored over the rationals.
But it has two real roots $sqrt{2}$ and $-sqrt{2}$ so if we factor it over the reals then it factors as $x^2 -2 = (x - sqrt{2})(x + sqrt{2})$.
A more illustrative example might be $x^3 - 2x^2 - x + 2$.
By the rational roots test and trial and error it has a root of $1$ ($1^3 -2 - 1 + 2 = 0$) so we can factor $x-1$ out of it and get:
$x^2 - 2x^2 - x + 2 = (x-1)(x^2 - 2)$ and $x^2 - 2$ can't be factored over the rationals so that is our final factorization.
But over the reals $x^2-2$ has two roots so it may be factored as $(x-1)(x+sqrt 2)(x - sqrt 2)$.
Finally consider $x^2 + 2 = 0$. This has no real roots at all so it can not be factored of the reals. (Nor over the rationals.)
And $x^3 - x^2 +2x - 2$ has only one real root. $1$. Which is a rational root. It can only be factored as far as $(x -1)(x^2 + 2)$ and that's as far as you can factor it over the rationals or over the reals.
answered Jan 5 at 22:53
fleabloodfleablood
72k22687
answered Jan 5 at 22:53
fleabloodfleablood
72k22687
answered Jan 5 at 22:53
fleabloodfleablood
72k22687
answered Jan 5 at 22:53
fleabloodfleablood
72k22687
72k22687
$begingroup$
On Michael Wang's comment concerning the factoring over the polynomial 2x^4-5x^3-4x^2+15x-6, I checked over his comment and found that his factoring the polynomial over the real numbers is incorrect. What would the correct solution be (add explanation if possible)?
$endgroup$
– James
Jan 5 at 23:16
$begingroup$
I believe the correct solution to the complete factoring over real numbers of 2x^4-5x^3-4x^2+15x-6 is (x-2)(2x-1)(x-root3)(x+root3). Correct me if I am wrong.
$endgroup$
– James
Jan 5 at 23:22
$begingroup$
If M.Wangs answer has an error its based on your work but it has no error. $x=2$ is a solution and we can factor $2x^4 - 5x^3 -4x^2+15x-6= 2x^4 - 4x^3 - x^3 +2x^2 -6x^2+12x + 3x -6 = (x-2)(2x^3-x^2 -6x+3)$ and we can factor $2x-1$ from that to get $(x-2)(2x-1)(2x-1)(x^2-3)$ and .... as you should be becoming familiar with $x^2 -3$ can not be factored over the rationals but it can be factor as $(x-sqrt3)(x+sqrt 3)$ so $(x-2)(2x-1)(2x-1)(x^2-3)$ is complete factorizing over the rationals and $(x-2)(2x-1)(2x-1)(x-sqrt3)(x+sqrt 3)$ is complete factorization over reals.
$endgroup$
– fleablood
Jan 6 at 4:08
add a comment |
$begingroup$
On Michael Wang's comment concerning the factoring over the polynomial 2x^4-5x^3-4x^2+15x-6, I checked over his comment and found that his factoring the polynomial over the real numbers is incorrect. What would the correct solution be (add explanation if possible)?
$endgroup$
– James
Jan 5 at 23:16
$begingroup$
I believe the correct solution to the complete factoring over real numbers of 2x^4-5x^3-4x^2+15x-6 is (x-2)(2x-1)(x-root3)(x+root3). Correct me if I am wrong.
$endgroup$
– James
Jan 5 at 23:22
$begingroup$
If M.Wangs answer has an error its based on your work but it has no error. $x=2$ is a solution and we can factor $2x^4 - 5x^3 -4x^2+15x-6= 2x^4 - 4x^3 - x^3 +2x^2 -6x^2+12x + 3x -6 = (x-2)(2x^3-x^2 -6x+3)$ and we can factor $2x-1$ from that to get $(x-2)(2x-1)(2x-1)(x^2-3)$ and .... as you should be becoming familiar with $x^2 -3$ can not be factored over the rationals but it can be factor as $(x-sqrt3)(x+sqrt 3)$ so $(x-2)(2x-1)(2x-1)(x^2-3)$ is complete factorizing over the rationals and $(x-2)(2x-1)(2x-1)(x-sqrt3)(x+sqrt 3)$ is complete factorization over reals.
$endgroup$
– fleablood
Jan 6 at 4:08
$begingroup$
On Michael Wang's comment concerning the factoring over the polynomial 2x^4-5x^3-4x^2+15x-6, I checked over his comment and found that his factoring the polynomial over the real numbers is incorrect. What would the correct solution be (add explanation if possible)?
$endgroup$
– James
Jan 5 at 23:16
$begingroup$
On Michael Wang's comment concerning the factoring over the polynomial 2x^4-5x^3-4x^2+15x-6, I checked over his comment and found that his factoring the polynomial over the real numbers is incorrect. What would the correct solution be (add explanation if possible)?
$endgroup$
– James
Jan 5 at 23:16
$begingroup$
I believe the correct solution to the complete factoring over real numbers of 2x^4-5x^3-4x^2+15x-6 is (x-2)(2x-1)(x-root3)(x+root3). Correct me if I am wrong.
$endgroup$
– James
Jan 5 at 23:22
$begingroup$
I believe the correct solution to the complete factoring over real numbers of 2x^4-5x^3-4x^2+15x-6 is (x-2)(2x-1)(x-root3)(x+root3). Correct me if I am wrong.
$endgroup$
– James
Jan 5 at 23:22
$begingroup$
If M.Wangs answer has an error its based on your work but it has no error. $x=2$ is a solution and we can factor $2x^4 - 5x^3 -4x^2+15x-6= 2x^4 - 4x^3 - x^3 +2x^2 -6x^2+12x + 3x -6 = (x-2)(2x^3-x^2 -6x+3)$ and we can factor $2x-1$ from that to get $(x-2)(2x-1)(2x-1)(x^2-3)$ and .... as you should be becoming familiar with $x^2 -3$ can not be factored over the rationals but it can be factor as $(x-sqrt3)(x+sqrt 3)$ so $(x-2)(2x-1)(2x-1)(x^2-3)$ is complete factorizing over the rationals and $(x-2)(2x-1)(2x-1)(x-sqrt3)(x+sqrt 3)$ is complete factorization over reals.
$endgroup$
– fleablood
Jan 6 at 4:08
$begingroup$
If M.Wangs answer has an error its based on your work but it has no error. $x=2$ is a solution and we can factor $2x^4 - 5x^3 -4x^2+15x-6= 2x^4 - 4x^3 - x^3 +2x^2 -6x^2+12x + 3x -6 = (x-2)(2x^3-x^2 -6x+3)$ and we can factor $2x-1$ from that to get $(x-2)(2x-1)(2x-1)(x^2-3)$ and .... as you should be becoming familiar with $x^2 -3$ can not be factored over the rationals but it can be factor as $(x-sqrt3)(x+sqrt 3)$ so $(x-2)(2x-1)(2x-1)(x^2-3)$ is complete factorizing over the rationals and $(x-2)(2x-1)(2x-1)(x-sqrt3)(x+sqrt 3)$ is complete factorization over reals.
$endgroup$
– fleablood
Jan 6 at 4:08
add a comment |
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$begingroup$
If you're asking about factoring polynomials then "factoring ovr the real numbers" means that the factors should be polynomials all of whose coefficients are real. If you're asking about something else, then you should say what you intend.
$endgroup$
– Andreas Blass
Jan 5 at 22:38
$begingroup$
Yes, I am asking about factoring polynomials. For instance, could you explain to me how to factor the following polynomial: 2x^4-5x^3-4x^2+15x-6
$endgroup$
– James
Jan 5 at 22:41
$begingroup$
Could you put "factor over the reals" in context? And what are you factoring? polynomials? numbers? It seems to me that the difference between of reals and integers is that you have any two reals $a,b$ and $b ne 0$ then you can always find a real $k$ so that $a = b*k$ so every non-zero number is a factor of every number whereas under integers to have $b|a$ is a special event that may not occur.
$endgroup$
– fleablood
Jan 5 at 22:42
$begingroup$
I am speaking in terms of polynomials.
$endgroup$
– James
Jan 5 at 22:48