Basic algebra: Prove that an n-degree polynomial is expressible as the product of n binomials.
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Using basic algebra, how does one prove that an n-degree polynomial is expressible as the product of $n$ binomials? Here I am allowing for binomials of the form $xb_i+alpha_i=x0+1.$
This is something I should know like the back of my hand, but I don't. The answer I am seeking is a high school-level demonstration. I am motivated by the following statement in Hermann Weyl's The Classical Groups:
A polynomial of degree $n$ has at most $n$ different zeros; this follows in the well-known way by proving that $fleft(xright)$ contains the factors $left(x — alpha_1right)left(x — alpha_2right)dots$ if $alpha_1,alpha_2,dots$ are distinct zeros.
I have sources which provide "advanced" proofs. I am seeking an intuitively satisfying proof that could be presented to a reasonably intelligent high school student with the expectation of understanding. I am more interested in the method of proof, than the proof itself.
algebra-precalculus proof-writing
asked Dec 7 '18 at 1:42
Steven HattonSteven Hatton
892418
$endgroup$
add a comment |
$begingroup$
Using basic algebra, how does one prove that an n-degree polynomial is expressible as the product of $n$ binomials? Here I am allowing for binomials of the form $xb_i+alpha_i=x0+1.$
This is something I should know like the back of my hand, but I don't. The answer I am seeking is a high school-level demonstration. I am motivated by the following statement in Hermann Weyl's The Classical Groups:
A polynomial of degree $n$ has at most $n$ different zeros; this follows in the well-known way by proving that $fleft(xright)$ contains the factors $left(x — alpha_1right)left(x — alpha_2right)dots$ if $alpha_1,alpha_2,dots$ are distinct zeros.
I have sources which provide "advanced" proofs. I am seeking an intuitively satisfying proof that could be presented to a reasonably intelligent high school student with the expectation of understanding. I am more interested in the method of proof, than the proof itself.
algebra-precalculus proof-writing
asked Dec 7 '18 at 1:42
Steven HattonSteven Hatton
892418
$endgroup$
$begingroup$
The intuitively satisfying version (IMHO) is indeed the algorithm of factorizing: Find a root, divide by the corresponding linear factor, see that you get remainder zero, and repeat.
$endgroup$
– T. Bongers
Dec 7 '18 at 1:45
$begingroup$
I believe I can provide a somewhat intuitive demonstration of this fact, but I do need to ask one question - are you familiar with complex numbers on some level? I ask, because the theorem "a polynomial of degree $n$ is the product of $n$ binomials" is analogous to the fundamental theorem of algebra, which is slightly different for real polynomials and complex polynomials. (Namely, real polynomials of degree $n$ have at most $n$ roots, and complex ones will have exactly $n$ roots.) It's a small but important nuance.
$endgroup$
– Eevee Trainer
Dec 7 '18 at 1:45
$begingroup$
For the record, my demonstration indeed would have been the factoring idea T. Bongers mentioned lol
$endgroup$
– Eevee Trainer
Dec 7 '18 at 1:46
$begingroup$
This is related to the fundamental theorem of algebra which has been proven many times, but never satisfactorily.
$endgroup$
– John L Winters
Dec 7 '18 at 2:32
add a comment |
$begingroup$
Using basic algebra, how does one prove that an n-degree polynomial is expressible as the product of $n$ binomials? Here I am allowing for binomials of the form $xb_i+alpha_i=x0+1.$
This is something I should know like the back of my hand, but I don't. The answer I am seeking is a high school-level demonstration. I am motivated by the following statement in Hermann Weyl's The Classical Groups:
A polynomial of degree $n$ has at most $n$ different zeros; this follows in the well-known way by proving that $fleft(xright)$ contains the factors $left(x — alpha_1right)left(x — alpha_2right)dots$ if $alpha_1,alpha_2,dots$ are distinct zeros.
I have sources which provide "advanced" proofs. I am seeking an intuitively satisfying proof that could be presented to a reasonably intelligent high school student with the expectation of understanding. I am more interested in the method of proof, than the proof itself.
algebra-precalculus proof-writing
asked Dec 7 '18 at 1:42
Steven HattonSteven Hatton
892418
$endgroup$
Using basic algebra, how does one prove that an n-degree polynomial is expressible as the product of $n$ binomials? Here I am allowing for binomials of the form $xb_i+alpha_i=x0+1.$
This is something I should know like the back of my hand, but I don't. The answer I am seeking is a high school-level demonstration. I am motivated by the following statement in Hermann Weyl's The Classical Groups:
A polynomial of degree $n$ has at most $n$ different zeros; this follows in the well-known way by proving that $fleft(xright)$ contains the factors $left(x — alpha_1right)left(x — alpha_2right)dots$ if $alpha_1,alpha_2,dots$ are distinct zeros.
I have sources which provide "advanced" proofs. I am seeking an intuitively satisfying proof that could be presented to a reasonably intelligent high school student with the expectation of understanding. I am more interested in the method of proof, than the proof itself.
algebra-precalculus proof-writing
algebra-precalculus proof-writing
asked Dec 7 '18 at 1:42
Steven HattonSteven Hatton
892418
asked Dec 7 '18 at 1:42
Steven HattonSteven Hatton
892418
edited Dec 7 '18 at 2:33
Steven Hatton
asked Dec 7 '18 at 1:42
Steven HattonSteven Hatton
892418
asked Dec 7 '18 at 1:42
Steven HattonSteven Hatton
892418
asked Dec 7 '18 at 1:42
Steven HattonSteven Hatton
892418
892418
$begingroup$
The intuitively satisfying version (IMHO) is indeed the algorithm of factorizing: Find a root, divide by the corresponding linear factor, see that you get remainder zero, and repeat.
$endgroup$
– T. Bongers
Dec 7 '18 at 1:45
$begingroup$
I believe I can provide a somewhat intuitive demonstration of this fact, but I do need to ask one question - are you familiar with complex numbers on some level? I ask, because the theorem "a polynomial of degree $n$ is the product of $n$ binomials" is analogous to the fundamental theorem of algebra, which is slightly different for real polynomials and complex polynomials. (Namely, real polynomials of degree $n$ have at most $n$ roots, and complex ones will have exactly $n$ roots.) It's a small but important nuance.
$endgroup$
– Eevee Trainer
Dec 7 '18 at 1:45
$begingroup$
For the record, my demonstration indeed would have been the factoring idea T. Bongers mentioned lol
$endgroup$
– Eevee Trainer
Dec 7 '18 at 1:46
$begingroup$
This is related to the fundamental theorem of algebra which has been proven many times, but never satisfactorily.
$endgroup$
– John L Winters
Dec 7 '18 at 2:32
add a comment |
$begingroup$
The intuitively satisfying version (IMHO) is indeed the algorithm of factorizing: Find a root, divide by the corresponding linear factor, see that you get remainder zero, and repeat.
$endgroup$
– T. Bongers
Dec 7 '18 at 1:45
$begingroup$
I believe I can provide a somewhat intuitive demonstration of this fact, but I do need to ask one question - are you familiar with complex numbers on some level? I ask, because the theorem "a polynomial of degree $n$ is the product of $n$ binomials" is analogous to the fundamental theorem of algebra, which is slightly different for real polynomials and complex polynomials. (Namely, real polynomials of degree $n$ have at most $n$ roots, and complex ones will have exactly $n$ roots.) It's a small but important nuance.
$endgroup$
– Eevee Trainer
Dec 7 '18 at 1:45
$begingroup$
For the record, my demonstration indeed would have been the factoring idea T. Bongers mentioned lol
$endgroup$
– Eevee Trainer
Dec 7 '18 at 1:46
$begingroup$
This is related to the fundamental theorem of algebra which has been proven many times, but never satisfactorily.
$endgroup$
– John L Winters
Dec 7 '18 at 2:32
$begingroup$
The intuitively satisfying version (IMHO) is indeed the algorithm of factorizing: Find a root, divide by the corresponding linear factor, see that you get remainder zero, and repeat.
$endgroup$
– T. Bongers
Dec 7 '18 at 1:45
$begingroup$
The intuitively satisfying version (IMHO) is indeed the algorithm of factorizing: Find a root, divide by the corresponding linear factor, see that you get remainder zero, and repeat.
$endgroup$
– T. Bongers
Dec 7 '18 at 1:45
$begingroup$
I believe I can provide a somewhat intuitive demonstration of this fact, but I do need to ask one question - are you familiar with complex numbers on some level? I ask, because the theorem "a polynomial of degree $n$ is the product of $n$ binomials" is analogous to the fundamental theorem of algebra, which is slightly different for real polynomials and complex polynomials. (Namely, real polynomials of degree $n$ have at most $n$ roots, and complex ones will have exactly $n$ roots.) It's a small but important nuance.
$endgroup$
– Eevee Trainer
Dec 7 '18 at 1:45
$begingroup$
I believe I can provide a somewhat intuitive demonstration of this fact, but I do need to ask one question - are you familiar with complex numbers on some level? I ask, because the theorem "a polynomial of degree $n$ is the product of $n$ binomials" is analogous to the fundamental theorem of algebra, which is slightly different for real polynomials and complex polynomials. (Namely, real polynomials of degree $n$ have at most $n$ roots, and complex ones will have exactly $n$ roots.) It's a small but important nuance.
$endgroup$
– Eevee Trainer
Dec 7 '18 at 1:45
$begingroup$
For the record, my demonstration indeed would have been the factoring idea T. Bongers mentioned lol
$endgroup$
– Eevee Trainer
Dec 7 '18 at 1:46
$begingroup$
For the record, my demonstration indeed would have been the factoring idea T. Bongers mentioned lol
$endgroup$
– Eevee Trainer
Dec 7 '18 at 1:46
$begingroup$
This is related to the fundamental theorem of algebra which has been proven many times, but never satisfactorily.
$endgroup$
– John L Winters
Dec 7 '18 at 2:32
$begingroup$
This is related to the fundamental theorem of algebra which has been proven many times, but never satisfactorily.
$endgroup$
– John L Winters
Dec 7 '18 at 2:32
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A polynomial of degree $n$ has at most $n$ different zeros; this follows in the well-known way by proving that $fleft(xright)$ contains the factors $left(x — alpha_1right)left(x — alpha_2right)dots$ if $alpha_1,alpha_2,dots$ are distinct zeros.
This statement does not claim that a polynomial of degree $n$ can be expressed as the product of $n$ binomials.
In fact, the claim in the title of your question is false in real algebra, although it is true in complex algebra.
What the claim in the book actually asserts is that if the numbers
$alpha_1,$ $alpha_2, ldots,$ $alpha_k$ are $k$ distinct zeros of a polynomial $f(x),$ then
$$f(x) = (x - alpha_1)(x - alpha_2)cdots(x - alpha_k)g(x) tag1$$
where $g(x)$ is a polynomial.
It never says that so many zeros necessarily exist.
A real polynomial of any even degree may have no real zeros at all. A real polynomial of any odd degree may have just one real zero and no others. In general, a polynomial $f(x)$ of degree $n$ can have fewer than $n$ zeros, but it cannot have more than $n$ zeros.
If it had more than $n$ zeros, the factorization in Equation $(1)$ would show that $f(x)$ was a polynomial of degree greater than $n,$ a contradiction.
answered Dec 7 '18 at 1:53
David KDavid K
53.8k342116
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add a comment |
$begingroup$
I don't think you will have a basic algebraic proof, at a high school level, because you require complex numbers to even deal with $x^2+1$. Even this supposedly simple proof requires calculus.
answered Dec 7 '18 at 1:46
RandomMathGuyRandomMathGuy
1192
$endgroup$
$begingroup$
I am reaching back more decades than I care to admit, but I vaguely recall that a demonstration was giving by my high school teacher. We did learn the basics of complex numbers, so the proof may have involved complex numbers. I believe explicit reference to complex numbers might be avoided by either expressing the problem in the language of linear algebra, or using the language of abstract algebra. Note that I have refined the proposition to "...n or fewer binomials."
$endgroup$
– Steven Hatton
Dec 7 '18 at 2:28
$begingroup$
@StevenHatton I'm sure that really advanced high school students can follow the link I posted, but I am not personally comfortable calling that a "high-school level" proof. Maybe my own experience with students taking calculus in college has soured me to the American public education system, but truthfully I don't think you will find an average high schooler that will follow. On a different note, the link below appears to shed some light on the state of the art; some amount of analysis or topology is required for the proof. mathoverflow.net/a/25045
$endgroup$
– RandomMathGuy
Dec 7 '18 at 2:32
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
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votes
$begingroup$
A polynomial of degree $n$ has at most $n$ different zeros; this follows in the well-known way by proving that $fleft(xright)$ contains the factors $left(x — alpha_1right)left(x — alpha_2right)dots$ if $alpha_1,alpha_2,dots$ are distinct zeros.
This statement does not claim that a polynomial of degree $n$ can be expressed as the product of $n$ binomials.
In fact, the claim in the title of your question is false in real algebra, although it is true in complex algebra.
What the claim in the book actually asserts is that if the numbers
$alpha_1,$ $alpha_2, ldots,$ $alpha_k$ are $k$ distinct zeros of a polynomial $f(x),$ then
$$f(x) = (x - alpha_1)(x - alpha_2)cdots(x - alpha_k)g(x) tag1$$
where $g(x)$ is a polynomial.
It never says that so many zeros necessarily exist.
A real polynomial of any even degree may have no real zeros at all. A real polynomial of any odd degree may have just one real zero and no others. In general, a polynomial $f(x)$ of degree $n$ can have fewer than $n$ zeros, but it cannot have more than $n$ zeros.
If it had more than $n$ zeros, the factorization in Equation $(1)$ would show that $f(x)$ was a polynomial of degree greater than $n,$ a contradiction.
answered Dec 7 '18 at 1:53
David KDavid K
53.8k342116
$endgroup$
add a comment |
$begingroup$
A polynomial of degree $n$ has at most $n$ different zeros; this follows in the well-known way by proving that $fleft(xright)$ contains the factors $left(x — alpha_1right)left(x — alpha_2right)dots$ if $alpha_1,alpha_2,dots$ are distinct zeros.
This statement does not claim that a polynomial of degree $n$ can be expressed as the product of $n$ binomials.
In fact, the claim in the title of your question is false in real algebra, although it is true in complex algebra.
What the claim in the book actually asserts is that if the numbers
$alpha_1,$ $alpha_2, ldots,$ $alpha_k$ are $k$ distinct zeros of a polynomial $f(x),$ then
$$f(x) = (x - alpha_1)(x - alpha_2)cdots(x - alpha_k)g(x) tag1$$
where $g(x)$ is a polynomial.
It never says that so many zeros necessarily exist.
A real polynomial of any even degree may have no real zeros at all. A real polynomial of any odd degree may have just one real zero and no others. In general, a polynomial $f(x)$ of degree $n$ can have fewer than $n$ zeros, but it cannot have more than $n$ zeros.
If it had more than $n$ zeros, the factorization in Equation $(1)$ would show that $f(x)$ was a polynomial of degree greater than $n,$ a contradiction.
answered Dec 7 '18 at 1:53
David KDavid K
53.8k342116
$endgroup$
add a comment |
$begingroup$
A polynomial of degree $n$ has at most $n$ different zeros; this follows in the well-known way by proving that $fleft(xright)$ contains the factors $left(x — alpha_1right)left(x — alpha_2right)dots$ if $alpha_1,alpha_2,dots$ are distinct zeros.
This statement does not claim that a polynomial of degree $n$ can be expressed as the product of $n$ binomials.
In fact, the claim in the title of your question is false in real algebra, although it is true in complex algebra.
What the claim in the book actually asserts is that if the numbers
$alpha_1,$ $alpha_2, ldots,$ $alpha_k$ are $k$ distinct zeros of a polynomial $f(x),$ then
$$f(x) = (x - alpha_1)(x - alpha_2)cdots(x - alpha_k)g(x) tag1$$
where $g(x)$ is a polynomial.
It never says that so many zeros necessarily exist.
A real polynomial of any even degree may have no real zeros at all. A real polynomial of any odd degree may have just one real zero and no others. In general, a polynomial $f(x)$ of degree $n$ can have fewer than $n$ zeros, but it cannot have more than $n$ zeros.
If it had more than $n$ zeros, the factorization in Equation $(1)$ would show that $f(x)$ was a polynomial of degree greater than $n,$ a contradiction.
answered Dec 7 '18 at 1:53
David KDavid K
53.8k342116
$endgroup$
A polynomial of degree $n$ has at most $n$ different zeros; this follows in the well-known way by proving that $fleft(xright)$ contains the factors $left(x — alpha_1right)left(x — alpha_2right)dots$ if $alpha_1,alpha_2,dots$ are distinct zeros.
This statement does not claim that a polynomial of degree $n$ can be expressed as the product of $n$ binomials.
In fact, the claim in the title of your question is false in real algebra, although it is true in complex algebra.
What the claim in the book actually asserts is that if the numbers
$alpha_1,$ $alpha_2, ldots,$ $alpha_k$ are $k$ distinct zeros of a polynomial $f(x),$ then
$$f(x) = (x - alpha_1)(x - alpha_2)cdots(x - alpha_k)g(x) tag1$$
where $g(x)$ is a polynomial.
It never says that so many zeros necessarily exist.
A real polynomial of any even degree may have no real zeros at all. A real polynomial of any odd degree may have just one real zero and no others. In general, a polynomial $f(x)$ of degree $n$ can have fewer than $n$ zeros, but it cannot have more than $n$ zeros.
If it had more than $n$ zeros, the factorization in Equation $(1)$ would show that $f(x)$ was a polynomial of degree greater than $n,$ a contradiction.
answered Dec 7 '18 at 1:53
David KDavid K
53.8k342116
edited Dec 7 '18 at 1:58
answered Dec 7 '18 at 1:53
David KDavid K
53.8k342116
answered Dec 7 '18 at 1:53
David KDavid K
53.8k342116
answered Dec 7 '18 at 1:53
David KDavid K
53.8k342116
53.8k342116
add a comment |
add a comment |
$begingroup$
I don't think you will have a basic algebraic proof, at a high school level, because you require complex numbers to even deal with $x^2+1$. Even this supposedly simple proof requires calculus.
answered Dec 7 '18 at 1:46
RandomMathGuyRandomMathGuy
1192
$endgroup$
$begingroup$
I am reaching back more decades than I care to admit, but I vaguely recall that a demonstration was giving by my high school teacher. We did learn the basics of complex numbers, so the proof may have involved complex numbers. I believe explicit reference to complex numbers might be avoided by either expressing the problem in the language of linear algebra, or using the language of abstract algebra. Note that I have refined the proposition to "...n or fewer binomials."
$endgroup$
– Steven Hatton
Dec 7 '18 at 2:28
$begingroup$
@StevenHatton I'm sure that really advanced high school students can follow the link I posted, but I am not personally comfortable calling that a "high-school level" proof. Maybe my own experience with students taking calculus in college has soured me to the American public education system, but truthfully I don't think you will find an average high schooler that will follow. On a different note, the link below appears to shed some light on the state of the art; some amount of analysis or topology is required for the proof. mathoverflow.net/a/25045
$endgroup$
– RandomMathGuy
Dec 7 '18 at 2:32
add a comment |
$begingroup$
I don't think you will have a basic algebraic proof, at a high school level, because you require complex numbers to even deal with $x^2+1$. Even this supposedly simple proof requires calculus.
answered Dec 7 '18 at 1:46
RandomMathGuyRandomMathGuy
1192
$endgroup$
$begingroup$
I am reaching back more decades than I care to admit, but I vaguely recall that a demonstration was giving by my high school teacher. We did learn the basics of complex numbers, so the proof may have involved complex numbers. I believe explicit reference to complex numbers might be avoided by either expressing the problem in the language of linear algebra, or using the language of abstract algebra. Note that I have refined the proposition to "...n or fewer binomials."
$endgroup$
– Steven Hatton
Dec 7 '18 at 2:28
$begingroup$
@StevenHatton I'm sure that really advanced high school students can follow the link I posted, but I am not personally comfortable calling that a "high-school level" proof. Maybe my own experience with students taking calculus in college has soured me to the American public education system, but truthfully I don't think you will find an average high schooler that will follow. On a different note, the link below appears to shed some light on the state of the art; some amount of analysis or topology is required for the proof. mathoverflow.net/a/25045
$endgroup$
– RandomMathGuy
Dec 7 '18 at 2:32
add a comment |
$begingroup$
I don't think you will have a basic algebraic proof, at a high school level, because you require complex numbers to even deal with $x^2+1$. Even this supposedly simple proof requires calculus.
answered Dec 7 '18 at 1:46
RandomMathGuyRandomMathGuy
1192
$endgroup$
I don't think you will have a basic algebraic proof, at a high school level, because you require complex numbers to even deal with $x^2+1$. Even this supposedly simple proof requires calculus.
answered Dec 7 '18 at 1:46
RandomMathGuyRandomMathGuy
1192
answered Dec 7 '18 at 1:46
RandomMathGuyRandomMathGuy
1192
answered Dec 7 '18 at 1:46
RandomMathGuyRandomMathGuy
1192
answered Dec 7 '18 at 1:46
RandomMathGuyRandomMathGuy
1192
1192
$begingroup$
I am reaching back more decades than I care to admit, but I vaguely recall that a demonstration was giving by my high school teacher. We did learn the basics of complex numbers, so the proof may have involved complex numbers. I believe explicit reference to complex numbers might be avoided by either expressing the problem in the language of linear algebra, or using the language of abstract algebra. Note that I have refined the proposition to "...n or fewer binomials."
$endgroup$
– Steven Hatton
Dec 7 '18 at 2:28
$begingroup$
@StevenHatton I'm sure that really advanced high school students can follow the link I posted, but I am not personally comfortable calling that a "high-school level" proof. Maybe my own experience with students taking calculus in college has soured me to the American public education system, but truthfully I don't think you will find an average high schooler that will follow. On a different note, the link below appears to shed some light on the state of the art; some amount of analysis or topology is required for the proof. mathoverflow.net/a/25045
$endgroup$
– RandomMathGuy
Dec 7 '18 at 2:32
add a comment |
$begingroup$
I am reaching back more decades than I care to admit, but I vaguely recall that a demonstration was giving by my high school teacher. We did learn the basics of complex numbers, so the proof may have involved complex numbers. I believe explicit reference to complex numbers might be avoided by either expressing the problem in the language of linear algebra, or using the language of abstract algebra. Note that I have refined the proposition to "...n or fewer binomials."
$endgroup$
– Steven Hatton
Dec 7 '18 at 2:28
$begingroup$
@StevenHatton I'm sure that really advanced high school students can follow the link I posted, but I am not personally comfortable calling that a "high-school level" proof. Maybe my own experience with students taking calculus in college has soured me to the American public education system, but truthfully I don't think you will find an average high schooler that will follow. On a different note, the link below appears to shed some light on the state of the art; some amount of analysis or topology is required for the proof. mathoverflow.net/a/25045
$endgroup$
– RandomMathGuy
Dec 7 '18 at 2:32
$begingroup$
I am reaching back more decades than I care to admit, but I vaguely recall that a demonstration was giving by my high school teacher. We did learn the basics of complex numbers, so the proof may have involved complex numbers. I believe explicit reference to complex numbers might be avoided by either expressing the problem in the language of linear algebra, or using the language of abstract algebra. Note that I have refined the proposition to "...n or fewer binomials."
$endgroup$
– Steven Hatton
Dec 7 '18 at 2:28
$begingroup$
I am reaching back more decades than I care to admit, but I vaguely recall that a demonstration was giving by my high school teacher. We did learn the basics of complex numbers, so the proof may have involved complex numbers. I believe explicit reference to complex numbers might be avoided by either expressing the problem in the language of linear algebra, or using the language of abstract algebra. Note that I have refined the proposition to "...n or fewer binomials."
$endgroup$
– Steven Hatton
Dec 7 '18 at 2:28
$begingroup$
@StevenHatton I'm sure that really advanced high school students can follow the link I posted, but I am not personally comfortable calling that a "high-school level" proof. Maybe my own experience with students taking calculus in college has soured me to the American public education system, but truthfully I don't think you will find an average high schooler that will follow. On a different note, the link below appears to shed some light on the state of the art; some amount of analysis or topology is required for the proof. mathoverflow.net/a/25045
$endgroup$
– RandomMathGuy
Dec 7 '18 at 2:32
$begingroup$
@StevenHatton I'm sure that really advanced high school students can follow the link I posted, but I am not personally comfortable calling that a "high-school level" proof. Maybe my own experience with students taking calculus in college has soured me to the American public education system, but truthfully I don't think you will find an average high schooler that will follow. On a different note, the link below appears to shed some light on the state of the art; some amount of analysis or topology is required for the proof. mathoverflow.net/a/25045
$endgroup$
– RandomMathGuy
Dec 7 '18 at 2:32
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The intuitively satisfying version (IMHO) is indeed the algorithm of factorizing: Find a root, divide by the corresponding linear factor, see that you get remainder zero, and repeat.
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– T. Bongers
Dec 7 '18 at 1:45
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I believe I can provide a somewhat intuitive demonstration of this fact, but I do need to ask one question - are you familiar with complex numbers on some level? I ask, because the theorem "a polynomial of degree $n$ is the product of $n$ binomials" is analogous to the fundamental theorem of algebra, which is slightly different for real polynomials and complex polynomials. (Namely, real polynomials of degree $n$ have at most $n$ roots, and complex ones will have exactly $n$ roots.) It's a small but important nuance.
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– Eevee Trainer
Dec 7 '18 at 1:45
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For the record, my demonstration indeed would have been the factoring idea T. Bongers mentioned lol
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– Eevee Trainer
Dec 7 '18 at 1:46
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This is related to the fundamental theorem of algebra which has been proven many times, but never satisfactorily.
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– John L Winters
Dec 7 '18 at 2:32